Proof step 1 from proof pconttr-lifted (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  lifted(trlp)(trf, βf), pmon(trlp)(trf), ¬ pcont(lift(trlp))(βf) ⊦

Rule argument

Additional information

Proof step 2 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ (# σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 ∧ lift(trlp)(σ1, α1) ∧ lift(trlp)(σ2, α2) → βf(σ1, α1) ⊑ βf(σ2, α2)) ∧ ppbd(lift(trlp))(βf)

Rule argument

Additional information

Used simplifier rules: montrl-relative-def: ⊦ pmon(trlp)(trf) ↔ (∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2)); liftedtr-def: ⊦ lifted(trlp)(trf, βf) ↔ (∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α)); conttr-relative-def: ⊦ pcont(trp)(βf) ↔ pmon(trp)(βf) ∧ ppbd(trp)(βf); montr-relative-def: ⊦ pmon(trp)(βf) ↔ (∀ σ1, α1, σ2, α2. # σ1 = # α1 + ⌜1⌝ ∧ # σ2 = # α2 + ⌜1⌝ ∧ σ1 ⊑ σ2 ∧ α1 ⊑ α2 ∧ trp(σ1, α1) ∧ trp(σ2, α2) → βf(σ1, α1) ⊑ βf(σ2, α2));

Proof step 3 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

  # σ1 = # α1 + ⌜1⌝ , # σ2 = # α2 + ⌜1⌝ , lift(trlp)(σ1, α1), lift(trlp)(σ2, α2), σ1 ⊑ σ2, α1 ⊑ α2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

• Proof step 55:

  ¬ ppbd(lift(trlp))(βf), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 4 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 5:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , lift(trlp)(σ2, α2), lift(trlp)(σ1, α1), α1 ⊑ α2, σ1 ⊑ σ2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), lift(trlp)(σ2, α2) → lift(trf)(σ2, α2) = βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used terms: σ2, α2

Proof step 5 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  # σ1 = # α1 + ⌜1⌝ , # σ2 = # α2 + ⌜1⌝ , lift(trf)(σ2, α2) = βf(σ2, α2), lift(trlp)(σ1, α1), lift(trlp)(σ2, α2), σ1 ⊑ σ2, α1 ⊑ α2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 6 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 7:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , lift(trf)(σ2, α2) = βf(σ2, α2), lift(trlp)(σ2, α2), lift(trlp)(σ1, α1), α1 ⊑ α2, σ1 ⊑ σ2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), lift(trlp)(σ1, α1) → lift(trf)(σ1, α1) = βf(σ1, α1), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used terms: σ1, α1

Proof step 7 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  # σ1 = # α1 + ⌜1⌝ , # σ2 = # α2 + ⌜1⌝ , lift(trf)(σ2, α2) = βf(σ2, α2), lift(trf)(σ1, α1) = βf(σ1, α1), lift(trlp)(σ1, α1), lift(trlp)(σ2, α2), σ1 ⊑ σ2, α1 ⊑ α2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Proof step 8 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 9:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , lift(trf)(σ1, α1) = βf(σ1, α1), lift(trf)(σ2, α2) = βf(σ2, α2), σ1 = ⌜sl⌝ , lift(trlp)(σ2, α2), lift(trlp)(σ1, α1), α1 ⊑ α2, σ1 ⊑ σ2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

• Proof step 49:

  # σ2 = # α2 + ⌜1⌝ , # σ1 = # α1 + ⌜1⌝ , lift(trf)(σ1, α1) = βf(σ1, α1), lift(trf)(σ2, α2) = βf(σ2, α2), σ1 = ⌜sf⌝ , lift(trlp)(σ2, α2), lift(trlp)(σ1, α1), α1 ⊑ α2, σ1 ⊑ σ2, ¬ βf(σ1, α1) ⊑ βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 9 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  # σ2 = # α2 + ⌜1⌝ , lift(trf)(σ2, α2) = βf(σ2, α2), lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), lift(trlp)(σ2, α2), ⌜sl⌝ ⊑ σ2, α1 ⊑ α2, ¬ βf(⌜sl⌝ , α1) ⊑ βf(σ2, α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 10 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 11:

  # σ2 = # α2 + ⌜1⌝ , lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lift(trf)(σ2, α2) = βf(σ2, α2), σ2 = ⌜sl0⌝ , ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(σ2, α2), lift(trlp)(⌜sl⌝ , α1), α1 ⊑ α2, ⌜sl⌝ ⊑ σ2, ¬ βf(⌜sl⌝ , α1) ⊑ βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

• Proof step 17:

  # σ2 = # α2 + ⌜1⌝ , lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lift(trf)(σ2, α2) = βf(σ2, α2), σ2 = ⌜sf⌝ , ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(σ2, α2), lift(trlp)(⌜sl⌝ , α1), α1 ⊑ α2, ⌜sl⌝ ⊑ σ2, ¬ βf(⌜sl⌝ , α1) ⊑ βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 11 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  lift(trf)(⌜sl0⌝ , α2) = βf(⌜sl0⌝ , α2), lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), ⌜# sl⌝ = # α1 + ⌜1⌝ , ⌜# sl0⌝ = # α2 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), lift(trlp)(⌜sl0⌝ , α2), α1 ⊑ α2, sl ⊑ sl0, ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sl0⌝ , α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 12 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 13:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lift(trf)(⌜sl0⌝ , α2) = βf(⌜sl0⌝ , α2), α2 = ⌜al⌝ , ⌜# sl0⌝ = # α2 + ⌜1⌝ , ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl0⌝ , α2), lift(trlp)(⌜sl⌝ , α1), sl ⊑ sl0, α1 ⊑ α2, ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sl0⌝ , α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 13 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  # al + 1 = # sl0, lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), ⌜# sl⌝ = # α1 + ⌜1⌝ , ⌜trf(sl0, al)⌝ = βf(⌜sl0⌝ , ⌜al⌝ ), lift(trlp)(⌜sl⌝ , α1), trlp(sl0, al), α1 ⊑ ⌜al⌝ , sl ⊑ sl0, ¬ βf(⌜sl⌝ , α1) ⊑ ⌜trf(sl0, al)⌝ , ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 14 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 15:

  # al + 1 = # sl0, lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), α1 = ⌜al0⌝ , ⌜trf(sl0, al)⌝ = βf(⌜sl0⌝ , ⌜al⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), trlp(sl0, al), sl ⊑ sl0, α1 ⊑ ⌜al⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ ⌜trf(sl0, al)⌝ , ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 15 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  # al0 + 1 = # sl, # al + 1 = # sl0, ⌜trf(sl, al0)⌝ = βf(⌜sl⌝ , ⌜al0⌝ ), ⌜trf(sl0, al)⌝ = βf(⌜sl0⌝ , ⌜al⌝ ), trlp(sl, al0), trlp(sl0, al), sl ⊑ sl0, al0 ⊑ al, ¬ trf(sl, al0) ⊑ trf(sl0, al), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 16 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: sl, al0, sl0, al

Proof step 17 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  # α2 = ∞, lift(trf)(⌜sf⌝ , α2) = βf(⌜sf⌝ , α2), lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), lift(trlp)(⌜sf⌝ , α2), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ α2, ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 18 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 19:

  # α2 = ∞, lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lift(trf)(⌜sf⌝ , α2) = βf(⌜sf⌝ , α2), α2 = ⌜af⌝ , ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sf⌝ , α2), lift(trlp)(⌜sl⌝ , α1), α1 ⊑ α2, ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 19 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n))); ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 20 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 21:

  ⊦ mon(σσ) ∧ σσ ≠ [] → (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ lim(σσ)) ∧ (∀ m. ¬ # lim(σσ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ ⌜lim(σσ).take(m)⌝ ⊑ σσ(n)))

• Proof step 22:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), α1 ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n))), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 21 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) → (∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ )) ∧ (∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n)))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ [] ≠ ⌜nf⌝ ; (s)

Proof step 23 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 24:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), α1 ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

• Proof step 36:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), α1 ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 24 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

• Proof step 26:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∃ n. ¬ ( ⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n + 1)) ⊦

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 25 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 26 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), α1 ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜nf⌝ (n) = nf(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 27 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), # sf.take(n + 1) = # af.take(n) + 1 ∧ # sf.take(2 + n) = # af.take(n + 1) + 1 ∧ af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + n) ∧ trlp(sf.take(n + 1), af.take(n)) ∧ trlp(sf.take(2 + n), af.take(n + 1)) → trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: sf.take(n + 1), af.take(n), sf.take(2 + n), af.take(n + 1)

Proof step 28 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 29:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), α1 ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + n) ∧ trlp(sf.take(n + 1), af.take(n)) ∧ trlp(sf.take(2 + n), af.take(n + 1))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ # nf.take(n) = n; (s)

Proof step 29 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 30:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ af.take(n) ⊑ af.take(n + 1), ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

• Proof step 32:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ sf.take(n + 1) ⊑ sf.take(2 + n), ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

• Proof step 34:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ trlp(sf.take(n + 1), af.take(n)), ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

• Proof step 35:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ trlp(sf.take(2 + n), af.take(n + 1)), ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Proof step 30 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

  ⊦ nf.take(m + 1) = nf.take(m) + nf(m) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: af, n

Proof step 31 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 32 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 33:

  ⊦ nf.take(m + 2) = nf.take(m + 1) + nf(m + 1) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: sf, n

Proof step 33 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 34 from proof pconttr-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 35 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n + 1

Proof step 36 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 37:

  lift(trf)(⌜sl⌝ , α1) = βf(⌜sl⌝ , α1), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), α1 = ⌜al⌝ , ⌜# sl⌝ = # α1 + ⌜1⌝ , lift(trlp)(⌜sl⌝ , α1), ⌜sl⌝ ⊑ ⌜sf⌝ , α1 ⊑ ⌜af⌝ , ¬ βf(⌜sl⌝ , α1) ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜m⌝ + ∞ = ∞; (s) add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 37 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  # al + 1 = # sl, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ; lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s)

Proof step 38 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 39:

  # al + 1 = # sl, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ⌜trf(sf.take(# al + 1), af.take(# al))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: # al

Proof step 39 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  # al + 1 = # sl, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜trf(sf.take(# sl), af.take(# al))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ⌜al⌝ ⊑ ⌜af⌝ , ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 40 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 41:

  # al0 = # al, # al + 1 = # sl, af = (⌜al0⌝ + ⌜nf0⌝ ).f, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜al⌝ ⊑ ⌜af⌝ , ⌜trf(sf.take(# sl), al0)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x0 = nf.take(n) ∧ nf0 = nf.drop(n) ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = n; (elim) Used terms: af, # al, al0, nf0

Proof step 41 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  # al0 = # al, # al + 1 = # sl, lim(⌜λ n. ⌜trf(sf.take(n + 1), (al0 + nf0).take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜sl⌝ ⊑ ⌜sf⌝ , ⌜trf(sf.take(# sl), al0)⌝ ⊑ βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), (al0 + nf0).take(n)), ∀ m. ¬ # βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ trf(sf.take(n + 1), (al0 + nf0).take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ (⌜x⌝ + ⌜nf⌝ ).f = x + nf; (s) ⊦ ⌜x + nf⌝ = ⌜x⌝ + ⌜nf⌝ ; (s)

Proof step 42 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply elim lemma

And obtained the following new goals

• Proof step 43:

  # al0 = # al, # sl0 = # sl, # al + 1 = # sl, lim(⌜λ n. ⌜trf(sf.take(n + 1), (al0 + nf0).take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), sf = (⌜sl0⌝ + ⌜sf0⌝ ).f, ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ⌜trf(sl0, al0)⌝ ⊑ βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜sl⌝ ⊑ ⌜sf⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ trf(sf.take(n + 1), (al0 + nf0).take(n))), ∀ n. trlp(sf.take(n + 1), (al0 + nf0).take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ x0 = nf.take(n) ∧ nf0 = nf.drop(n) ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = n; (elim) Used terms: sf, # sl, sl0, sf0

Proof step 43 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  # sl0 = # sl, # al0 = # al, # al + 1 = # sl, lim(⌜λ n. ⌜trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))⌝ ⌝ ) = βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ⌜sl⌝ ⊑ ⌜sl0⌝ + ⌜sf0⌝ , ⌜trf(sl0, al0)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp((sl0 + sf0).take(n + 1), (al0 + nf0).take(n)), ∀ m. ¬ # βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ (⌜x⌝ + ⌜nf⌝ ).f = x + nf; (s) ⊦ ⌜x + nf⌝ = ⌜x⌝ + ⌜nf⌝ ; (s)

Proof step 44 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

• Proof step 46:

  # al0 = # al, # sl0 = # sl, # al + 1 = # sl, lim(⌜λ n. ⌜trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))⌝ ⌝ ) = βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜trf(sl, al)⌝ = βf(⌜sl⌝ , ⌜al⌝ ), trlp(sl, al), ⌜trf(sl0, al0)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ¬ ⌜trf(sl, al)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ∀ m. ¬ # βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))), ∀ n. trlp((sl0 + sf0).take(n + 1), (al0 + nf0).take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∃ σ1. ⌜sl⌝ + σ1 = ⌜sl0⌝ + ⌜sf0⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sl⌝ , ⌜sl0⌝ + ⌜sf0⌝

Proof step 45 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 46 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  # al0 = # al, # al + 1 = # sl0, lim(⌜λ n. ⌜trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))⌝ ⌝ ) = βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜trf(sl0, al)⌝ = βf(⌜sl0⌝ , ⌜al⌝ ), trlp(sl0, al), ⌜trf(sl0, al0)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ⌜al⌝ ⊑ ⌜al0⌝ + ⌜nf0⌝ , ¬ ⌜trf(sl0, al)⌝ ⊑ βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ), ∀ m. ¬ # βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sl0⌝ + ⌜sf0⌝ , ⌜al0⌝ + ⌜nf0⌝ ).take(m) ⊑ trf((sl0 + sf0).take(n + 1), (al0 + nf0).take(n))), ∀ n. trlp((sl0 + sf0).take(n + 1), (al0 + nf0).take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream ⊦ # x = # y → (⌜x⌝ + σ = ⌜y⌝ + σ0 ↔ x = y ∧ σ = σ0); (s)

Proof step 47 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 48:

  ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = # y → (⌜x⌝ + σ = ⌜y⌝ + σ0 ↔ x = y ∧ σ = σ0); (s) append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream Used terms: ⌜al⌝ , ⌜al0⌝ + ⌜nf0⌝

Proof step 48 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 49 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  # σ2 = # α2 + ⌜1⌝ , # α1 = ∞, lift(trf)(σ2, α2) = βf(σ2, α2), lift(trf)(⌜sf⌝ , α1) = βf(⌜sf⌝ , α1), lift(trlp)(⌜sf⌝ , α1), lift(trlp)(σ2, α2), ⌜sf⌝ ⊑ σ2, α1 ⊑ α2, ¬ βf(⌜sf⌝ , α1) ⊑ βf(σ2, α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 50 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 51:

  # α1 = ∞, # σ2 = # α2 + ⌜1⌝ , lift(trf)(⌜sf⌝ , α1) = βf(⌜sf⌝ , α1), lift(trf)(σ2, α2) = βf(σ2, α2), σ2 = ⌜sf0⌝ , lift(trlp)(σ2, α2), lift(trlp)(⌜sf⌝ , α1), α1 ⊑ α2, ⌜sf⌝ ⊑ σ2, ¬ βf(⌜sf⌝ , α1) ⊑ βf(σ2, α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ¬ ⌜nf⌝ ⊑ ⌜x⌝ ; (s)

Proof step 51 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  # α2 = ∞, # α1 = ∞, lift(trf)(⌜sf0⌝ , α1) = βf(⌜sf0⌝ , α1), lift(trf)(⌜sf0⌝ , α2) = βf(⌜sf0⌝ , α2), lift(trlp)(⌜sf0⌝ , α2), lift(trlp)(⌜sf0⌝ , α1), α1 ⊑ α2, ¬ βf(⌜sf0⌝ , α1) ⊑ βf(⌜sf0⌝ , α2), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 52 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 53:

  # α1 = ∞, # α2 = ∞, lift(trf)(⌜sf0⌝ , α2) = βf(⌜sf0⌝ , α2), lift(trf)(⌜sf0⌝ , α1) = βf(⌜sf0⌝ , α1), α2 = ⌜af⌝ , lift(trlp)(⌜sf0⌝ , α1), lift(trlp)(⌜sf0⌝ , α2), α1 ⊑ α2, ¬ βf(⌜sf0⌝ , α1) ⊑ βf(⌜sf0⌝ , α2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 53 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 54:

  # α1 = ∞, lift(trf)(⌜sf0⌝ , α1) = βf(⌜sf0⌝ , α1), lim(⌜λ n. ⌜trf(sf0.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf0⌝ , ⌜af⌝ ), lift(trlp)(⌜sf0⌝ , α1), α1 ⊑ ⌜af⌝ , ¬ βf(⌜sf0⌝ , α1) ⊑ βf(⌜sf0⌝ , ⌜af⌝ ), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf0.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n))); ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 54 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n))); Stream[A] freely generated by ⌝, ⌝; ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ⌜nf⌝ ⊑ ⌜nf0⌝ ↔ nf = nf0; (s) ⊦ σ ⊑ σ; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 55 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 56:

  ⊦ ppbd(trp)(βf) ↔ (∀ sf, af, bl. trp(⌜sf⌝ , ⌜af⌝ ) → ((∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ ))

• Proof step 57:

  ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∃ sf, af, bl. ¬(lift(trlp)(⌜sf⌝ , ⌜af⌝ ) → ((∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ )) ⊦

Rule argument

Additional information

Used terms: lift(trlp), βf

Proof step 56 from proof pconttr-lifted (previous step) (back to tree)

Lemma pbdtr-relative-def

Proof step 57 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 58:

  ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n)));

Proof step 58 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 59:

  ⊦ mon(σσ) ∧ σσ ≠ [] → (∀ n. ⌜n⌝ < # σσ → σσ(n) ⊑ lim(σσ)) ∧ (∀ m. ¬ # lim(σσ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ ⌜lim(σσ).take(m)⌝ ⊑ σσ(n)))

• Proof step 60:

  mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) ∧ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ≠ [] → (∀ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. ⌜n⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ∧ ⌜lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m)⌝ ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n))), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 59 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 60 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) → (∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ )) ∧ (∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n)))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ [] ≠ ⌜nf⌝ ; (s)

Proof step 61 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 62:

  ¬ mon(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 74:

  ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 62 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 63:

  ⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ(n) ⊑ σσ(n + 1))

• Proof step 64:

  ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∃ n. ¬ ( ⌜n⌝ + ⌜1⌝ < # ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ → ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n) ⊑ ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ (n + 1)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: ⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝

Proof step 63 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 64 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 65:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ⌜nf⌝ (n) = nf(n); (s) add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf ⊦ ⌜n⌝ ≠ ∞; (s) ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s) ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s) ⊦ # ⌜nf⌝ = ∞; (s)

Proof step 65 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 66:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), # sf.take(n + 1) = # af.take(n) + 1 ∧ # sf.take(2 + n) = # af.take(n + 1) + 1 ∧ af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + n) ∧ trlp(sf.take(n + 1), af.take(n)) ∧ trlp(sf.take(2 + n), af.take(n + 1)) → trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Used terms: sf.take(n + 1), af.take(n), sf.take(2 + n), af.take(n + 1)

Proof step 66 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 67:

  ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , af.take(n) ⊑ af.take(n + 1) ∧ sf.take(n + 1) ⊑ sf.take(2 + n) ∧ trlp(sf.take(n + 1), af.take(n)) ∧ trlp(sf.take(2 + n), af.take(n + 1))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ # nf.take(n) = n; (s)

Proof step 67 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 68:

  ¬ af.take(n) ⊑ af.take(n + 1), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 70:

  ¬ sf.take(n + 1) ⊑ sf.take(2 + n), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 72:

  ¬ trlp(sf.take(n + 1), af.take(n)), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

• Proof step 73:

  ¬ trlp(sf.take(2 + n), af.take(n + 1)), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ trf(sf.take(2 + n), af.take(n + 1)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ (∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ) ↔ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 68 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 69:

  ⊦ nf.take(m + 1) = nf.take(m) + nf(m) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: af, n

Proof step 69 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 70 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 71:

  ⊦ nf.take(m + 2) = nf.take(m + 1) + nf(m + 1) '

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 + x2 ↔ x0 = x1); (s) a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: sf, n

Proof step 71 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 72 from proof pconttr-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 73 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n + 1

Proof step 74 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 75:

  ¬ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

• Proof step 102:

  βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦ ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝

Rule argument

Additional information

Proof step 75 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 76:

  ¬ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , lift(trlp)(⌜sf⌝ , ⌜af⌝ ) → lift(trf)(⌜sf⌝ , ⌜af⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used terms: ⌜sf⌝ , ⌜af⌝

Proof step 76 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 77:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), ¬ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n))); lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ );

Proof step 77 from proof pconttr-lifted (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 78:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , ¬ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

• Proof step 88:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = βf(⌜sf⌝ , ⌜af⌝ ), lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , ¬ βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ βf(⌜sf⌝ , ⌜af⌝ ), ∀ m. ¬ # βf(⌜sf⌝ , ⌜af⌝ ) < ⌜m⌝ → (∃ n. βf(⌜sf⌝ , ⌜af⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: Stream[A] freely generated by ⌝, ⌝;

Proof step 78 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 79:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , ¬ bl0 ⊑ bl, ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ m. ¬ # bl0 < m → (∃ n. bl0.take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ .take(n) = x.take(n); (s) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ # ⌜x⌝ = ⌜# x⌝ ; (s) ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 79 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 80:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , ¬ bl0 ⊑ bl, ¬ # bl0 < # bl0 → (∃ n. bl0.take(# bl0) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: # bl0

Proof step 80 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 81:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl0 ⊑ trf(sf.take(n + 1), af.take(n)), ¬ bl0 ⊑ bl, ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = n → x.take(n) = x; (s) ⊦ ¬ n < n; (s)

Proof step 81 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 82:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bl0⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bl0⌝ , bl0 ⊑ trf(sf.take(n + 1), af.take(n)), trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ¬ bl0 ⊑ bl, ∀ n. trf(sf.take(n + 1), af.take(n)) ⊑ bl0, ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: n

Proof step 82 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 83:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x0 = # x1 → (x0 ⊑ x1 ↔ x0 = x1); (s) ⊦ x ⊑ y → (y ⊑ x ↔ x = y); (s)

Proof step 83 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 84:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: n

Proof step 84 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 85:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, lift(trlp)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) → lift(trf)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝

Proof step 85 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 86:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, trlp(sf.take(n + 1), af.take(n)) → ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ;

Proof step 86 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 87:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜trf(sf.take(n + 1), af.take(n))⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ¬ trlp(sf.take(n + 1), af.take(n)), ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n0. trf(sf.take(n0 + 1), af.take(n0)) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 87 from proof pconttr-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 88 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 89:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ m. ∃ n. bf.take(m) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = nf.take(n); (s) ⊦ ¬ ∞ < n∞; (s) ⊦ # ⌜nf⌝ = ∞; (s) ⊦ ¬ ⌜nf⌝ ⊑ ⌜x⌝ ; (s)

Proof step 89 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 90:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∃ n. bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: # bl + 1

Proof step 90 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 91:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 91 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 92:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: n

Proof step 92 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 93:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), lift(trlp)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) → lift(trf)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝

Proof step 93 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 94:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , trlp(sf.take(n + 1), af.take(n)) → ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ;

Proof step 94 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 95:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ¬ trlp(sf.take(n + 1), af.take(n)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

• Proof step 96:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 95 from proof pconttr-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 96 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 97:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), trf(sf.take(n + 1), af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ bl, ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Used simplifier rules: prefix-trans: ⊦ x ⊑ y ∧ y ⊑ z → x ⊑ z; (forward) from specification list-last ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 97 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 98:

  ⊦ x ⊑ y ↔ (∃ z. x + z = y)

• Proof step 99:

  lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), trf(sf.take(n + 1), af.take(n)) ⊑ bl, bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ , ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∃ z. bf.take(# bl + 1) + z = bl ⊦

Rule argument

Additional information

Used terms: bf.take(# bl + 1), bl

Proof step 98 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 99 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 100:

  bf.take(# bl + 1) + bl0 = bl, lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) = ⌜bf⌝ , βf(⌜sf⌝ , ⌜af⌝ ) = ⌜bf⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), bf.take(# bl + 1) ⊑ trf(sf.take(n + 1), af.take(n)), trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ ⌜bf⌝ ⊦

Rule argument

Additional information

Proof step 100 from proof pconttr-lifted (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 101:

  ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x(n) = y(n))

Rule argument

Additional information

Used simplifier rules: ⊦ #(x + y) = # x + # y; (s) ⊦ # nf.take(n) = n; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ m = m + n ↔ n = 0; (s) ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: bf.take(# bl + 1) + bl0, bl

Proof step 101 from proof pconttr-lifted (previous step) (back to tree)

A spec-lemma

Proof step 102 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 103:

  βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 103 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 104:

  βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used terms: n

Proof step 104 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 105:

  ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , lift(trlp)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) → lift(trf)(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used terms: ⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝

Proof step 105 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 106:

  βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , trlp(sf.take(n + 1), af.take(n)) → ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ n. trlp(sf.take(n + 1), af.take(n)) ⊦

Rule argument

Additional information

Used simplifier rules: lift-trp-fin: ⊦ lift(trlp)(⌜sl⌝ , ⌜al⌝ ) ↔ trlp(sl, al); lifttr-fin: ⊦ lift(trf)(⌜sl⌝ , ⌜al⌝ ) = ⌜trf(sl, al)⌝ ;

Proof step 106 from proof pconttr-lifted (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 107:

  ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ trlp(sf.take(n + 1), af.take(n)), ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

• Proof step 108:

  ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ) ⊑ ⌜bl⌝ , ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Proof step 107 from proof pconttr-lifted (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n

Proof step 108 from proof pconttr-lifted (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 109:

  ⌜trf(sf.take(n + 1), af.take(n))⌝ = βf(⌜sf.take(n + 1)⌝ , ⌜af.take(n)⌝ ), ⌜trf(sf.take(n + 1), af.take(n))⌝ ⊑ lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ), βf(⌜sf⌝ , ⌜af⌝ ) ⊑ ⌜bl⌝ , ¬ trf(sf.take(n + 1), af.take(n)) ⊑ bl, ∀ n. trlp(sf.take(n + 1), af.take(n)), ∀ sl1, al1, sl2, al2. # sl1 = # al1 + 1 ∧ # sl2 = # al2 + 1 ∧ al1 ⊑ al2 ∧ sl1 ⊑ sl2 ∧ trlp(sl1, al1) ∧ trlp(sl2, al2) → trf(sl1, al1) ⊑ trf(sl2, al2), ∀ σ, α. lift(trlp)(σ, α) → lift(trf)(σ, α) = βf(σ, α), ∀ m. ¬ # lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ) < ⌜m⌝ → (∃ n. lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ).take(m) ⊑ trf(sf.take(n + 1), af.take(n))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s)

Proof step 109 from proof pconttr-lifted (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lift-trp-inf: ⊦ lift(trlp)(⌜sf⌝ , ⌜af⌝ ) ↔ (∀ n. trlp(sf.take(n + 1), af.take(n))); lifttr-inf: ⊦ lift(trf)(⌜sf⌝ , ⌜af⌝ ) = lim(⌜λ n. ⌜trf(sf.take(n + 1), af.take(n))⌝ ⌝ ); leq-trans: ⊦ σ ⊑ σ1 ∧ σ1 ⊑ σ2 → σ ⊑ σ2; (forward) from specification stream ⊦ ⌜x⌝ ⊑ ⌜y⌝ ↔ x ⊑ y; (s) Used terms: ⌜sf⌝ , ⌜af⌝