Proof step 1 from proof inv-get-lslen (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(⌜inv(nσ)(n)⌝ < # nσ ↔ true), ⌜n⌝ < leninv(nσ), stutf(nσ) ⊦

Rule argument

Additional information

Proof step 2 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜inv(nσ)(n)⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 3 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 4:

  ⊦ stutf(nσ) → inv(nσ) = (leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ .take(leninv(nσ).n))

• Proof step 5:

  ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜(leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ .take(leninv(nσ).n))(n) ⌝ < # nσ ⊦

Rule argument

Additional information

Used terms: nσ

Proof step 4 from proof inv-get-lslen (previous step) (back to tree)

Lemma inv-def

Proof step 5 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜(leninv(nσ) = ∞ ⊃ ⌜λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n⌝ ;⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(leninv(nσ).n)⌝ ) (n) ⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜nf⌝ .take(n) = ⌜nf.take(n)⌝ ; (s)

Proof step 6 from proof inv-get-lslen (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 7:

  leninv(nσ) ≠ ∞, ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(leninv(nσ).n)⌝ (n)⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: leninv-inf-strong: ⊦ stutf(nσ) → ( leninv(nσ) = ∞ ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ(m) < nσ(m0)) ∧ (∀ m. ∃ m0. m < nσ(m0))); ⊦ ⌜nf⌝ (n) = nf(n); (s) ⊦ ⌜m⌝ < ∞; (s)

Proof step 7 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  leninv(nσ) ≠ ∞, ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(leninv(nσ).n)(n)⌝ < # nσ ⊦

Rule argument

Additional information

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

Proof step 8 from proof inv-get-lslen (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 9:

  leninv(nσ) = ⌜n0⌝ , leninv(nσ) ≠ ∞, ⌜n⌝ < leninv(nσ), stutf(nσ), ¬ ⌜(λ m. max∞(λ n. ⌜n⌝ < # nσ ∧ nσ(n) = m).n).take(n0)(n)⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: leninv(nσ), n0

Proof step 9 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  leninv(nσ) = ⌜n0⌝ , n < n0, stutf(nσ), ¬ ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic m < n ⊦ nf.take(n)(m) = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 10 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  n∞ ≠ ∞ ⊦ ⌜n∞.n⌝ = n∞

• Proof step 12:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n < n0, stutf(nσ), ¬ ⌜max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n).n⌝ < # nσ ⊦

• Proof step 33:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) ≠ ∞, n < n0, stutf(nσ), ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) < # nσ ⊦

Rule argument

Additional information

Used terms: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n)

Proof step 11 from proof inv-get-lslen (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n < n0, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 13 from proof inv-get-lslen (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 14:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n0 = n1.+1, n < n0, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: Nat freely generated by 0, +1; ⊦ ¬ n < 0; (s)

Proof step 14 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  leninv(nσ) = ⌜n1⌝ + ⌜1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n ≤ n1, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m.+1 = m + 1; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 15 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ stutf(nσ) → (leninv(nσ) = ⌜n⌝ + ⌜1⌝ ↔ (# nσ = ∞ ⊃ ∃ m. nσ(m) = n ∧ (∀ m0. m < m0 → nσ(m0) = n + 1);nσ((# nσ).n - 1) = n))

• Proof step 17:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n ≤ n1, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ, (# nσ = ∞ ⊃ ∃ m. nσ(m) = n1 ∧ (∀ m0. m < m0 → nσ(m0) = n1 + 1);nσ((# nσ).n - 1) = n1) ⊦

Rule argument

Additional information

Used terms: n1, nσ

Proof step 16 from proof inv-get-lslen (previous step) (back to tree)

Lemma leninv-succ

Proof step 17 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, n ≤ n1, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ, (# nσ = ∞ ⊃ nσ(m) = n1 ∧ (∀ m0. m < m0 → nσ(m0) = nσ(m) + 1);nσ((# nσ).n - 1) = n1) ⊦

Rule argument

Additional information

Proof step 18 from proof inv-get-lslen (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 19:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, nσ((# nσ).n - 1) = n1, # nσ ≠ ∞, n ≤ n1, stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ < ∞; (s)

Proof step 19 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, # nσ ≠ ∞, n ≤ nσ((# nσ).n - 1), stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 20 from proof inv-get-lslen (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 21:

  # nσ = ⌜n1⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ∞, # nσ ≠ ∞, n ≤ nσ(n1 - 1), stutf(nσ), ¬ ⌜∞.n⌝ < # nσ ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # nσ, n1

Proof step 21 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  # nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ(n0) = n) = ∞, n ≤ nσ(n1 - 1), n1 ≤ ∞.n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s) ⊦ ⌜n⌝ ≠ ∞; (s)

Proof step 22 from proof inv-get-lslen (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 23:

  # nσ = ⌜n1⌝ , max∞(λ n0. n0 < n1 ∧ nσ(n0) = n) = ∞, n1 = m + 1, n1 ≠ 0, n ≤ nσ(m), n1 ≤ ∞.n, stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); ⊦ ¬ n < 0; (s) ⊦ # σ = ⌜0⌝ ↔ σ = []; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n1, m

Proof step 23 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  # nσ = ⌜m⌝ + ⌜1⌝ , max∞(λ n0. ¬ m < n0 ∧ nσ(n0) = n) = ∞, m < ∞.n, n ≤ nσ(m), stutf(nσ) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m + 1 ≠ 0; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 24 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

  ⊦ max∞(np) = ∞ ↔ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n))

• Proof step 26:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, n ≤ nσ(m), stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ(n1) = n) ∨ (∀ n0. ¬(¬ m < n0 ∧ nσ(n0) = n)) ⊦

Rule argument

Additional information

Used terms: λ n0. ¬ m < n0 ∧ nσ(n0) = n

Proof step 25 from proof inv-get-lslen (previous step) (back to tree)

Lemma maxinf-inf

Proof step 26 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, n ≤ nσ(m), stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ(n1) = n) ∨ (∀ n0. m < n0 ∨ nσ(n0) ≠ n) ⊦

Rule argument

Additional information

Proof step 27 from proof inv-get-lslen (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 28:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, n ≤ nσ(m), stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ(n1) = n ⊦

• Proof step 29:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, n ≤ nσ(m), stutf(nσ), ∀ n0. m < n0 ∨ nσ(n0) ≠ n ⊦

Rule argument

Additional information

Proof step 28 from proof inv-get-lslen (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m

Proof step 29 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 30:

  ⊦ stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n)

• Proof step 31:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, n ≤ nσ(m), stutf(nσ), ∀ n0. m < n0 ∨ nσ(n0) ≠ n, stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ(m) → (∃ m0. m0 ≤ m ∧ nσ(m0) = n) ⊦

Rule argument

Additional information

Used terms: nσ, m, n

Proof step 30 from proof inv-get-lslen (previous step) (back to tree)

Lemma stutf-alltaken

Proof step 31 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  # nσ = ⌜m⌝ + ⌜1⌝ , m < ∞.n, m0 < ∞.n, m0 ≤ m, nσ(m0) ≤ nσ(m), stutf(nσ), ∀ n0. m < n0 ∨ nσ(n0) ≠ nσ(m0) ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); (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

Proof step 32 from proof inv-get-lslen (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: m0

Proof step 33 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) ≠ ∞, n < n0, stutf(nσ), ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) < # nσ ⊦

Rule argument

Additional information

Proof step 34 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ n∞ = ∞ ↔ ¬(∃ n. n∞ = ⌜n⌝ )

• Proof step 36:

  leninv(nσ) = ⌜n0⌝ , n < n0, stutf(nσ), ¬ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) < # nσ, ∃ n0. max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n0⌝ ⊦

Rule argument

Additional information

Used terms: max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n)

Proof step 35 from proof inv-get-lslen (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof inv-get-lslen (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  leninv(nσ) = ⌜n0⌝ , max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n) = ⌜n1⌝ , n < n0, stutf(nσ), ¬ ⌜n1⌝ < # nσ ⊦

Rule argument

Additional information

Proof step 37 from proof inv-get-lslen (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

  ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

Rule argument

Additional information

Used terms: n1, λ n0. ⌜n0⌝ < # nσ ∧ nσ(n0) = n

Proof step 38 from proof inv-get-lslen (previous step) (back to tree)

Lemma maxinf-fin