Proof step 1 from proof stutf-alltakenmax (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ⌜m⌝ < # nσ, n < nσ(m), stutf(nσ) ⊦ ∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = n + 1

Rule argument

Additional information

Proof step 2 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  ⌜m⌝ < # nσ, n < nσ(m), stutf(nσ) ⊦ ∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 3 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule induction

And obtained the following new goals

• Proof step 4:

  ⌜m⌝ < # nσ, n < nσ(m), stutf(nσ), ∀ m0, nσ, n. m0 < m → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Used simplifier rules: <; (wellfounded) Used terms: m

Proof step 4 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ⌜m⌝ < # nσ, n < nσ(m), stutf(nσ), ∀ m0, nσ, n. m0 < m → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Proof step 5 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 6:

  m = m2.+1, ⌜m⌝ < # nσ, n < nσ(m), stutf(nσ), ∀ m0, nσ, n. m0 < m → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. m0 < m ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0; Nat freely generated by 0, +1; ⊦ ¬ n < 0; (s)

Proof step 6 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

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 < n + 1 ↔ ¬ n < m; (s) ⊦ m.+1 = m + 1; (s) ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)

Proof step 7 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 8:

  ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, ¬ m2 < m2 ∧ nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Used terms: m2

Proof step 8 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)

Proof step 9 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule apply lemma

And obtained the following new goals

• Proof step 10:

  ⊦ stutf(nσ) ↔ nσ ≠ [] ∧ nσ(0) = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1)

• Proof step 11:

  ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), stutf(nσ) ↔ nσ ≠ [] ∧ nσ(0) = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1), ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Used terms: nσ

Proof step 10 from proof stutf-alltakenmax (previous step) (back to tree)

Lemma stutf-def

Proof step 11 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Used simplifier rules: stutf-nonempty: ⊦ stutf(nσ) → (nσ = [] ↔ false); stutf-zero: ⊦ stutf(nσ) → nσ(0) = 0;

Proof step 12 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 13:

  nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ⌜m2⌝ + ⌜1⌝ < # nσ → nσ(m2 + 1) = nσ(m2) ∨ nσ(m2 + 1) = nσ(m2) + 1, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Used terms: m2

Proof step 13 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, nσ(m2 + 1) = nσ(m2) ∨ nσ(m2 + 1) = nσ(m2) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Proof step 14 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 15:

  nσ(m2 + 1) = nσ(m2), nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

• Proof step 19:

  nσ(m2 + 1) = nσ(m2) + 1, nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2 + 1), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1, nσ(m2) = n ∧ nσ(m2 + 1) = nσ(m2) + 1

Rule argument

Additional information

Proof step 15 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  nσ(m2 + 1) = nσ(m2), nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 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 = m + n ↔ n = 0; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 16 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule apply induction

And obtained the following new goals

• Proof step 17:

  nσ(m2 + 1) = nσ(m2), nσ ≠ [], ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2), stutf(nσ), ∃ m1. m1 < m2 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 ⊦ m.+1 = m + 1; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) Used terms: m2, nσ, n

Proof step 17 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  nσ(m1 + 1) = nσ(m1) + 1, nσ(m2 + 1) = nσ(m2), nσ ≠ [], m1 < m2, ⌜m2⌝ + ⌜1⌝ < # nσ, nσ(m1) < nσ(m2), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = nσ(m1) ∧ nσ(m0 + 1) = nσ(m0) + 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 + 1 ≤ n ↔ m < n; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat

Proof step 18 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s) Used terms: m1

Proof step 19 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  nσ(m2 + 1) = nσ(m2) + 1, nσ ≠ [], nσ(m2) ≠ n, ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

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 ≠ n → (m < n + 1 ↔ m < n); (s) ⊦ m + 1 ≠ 0; (s)

Proof step 20 from proof stutf-alltakenmax (previous step) (back to tree)

User applied rule apply induction

And obtained the following new goals

• Proof step 21:

  nσ(m2 + 1) = nσ(m2) + 1, nσ ≠ [], nσ(m2) ≠ n, ⌜m2⌝ + ⌜1⌝ < # nσ, n < nσ(m2), stutf(nσ), ∃ m1. m1 < m2 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = n ∧ nσ(m0 + 1) = nσ(m0) + 1

Rule argument

Additional information

Used simplifier rules: lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic 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 ⊦ m.+1 = m + 1; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) Used terms: m2, nσ, n

Proof step 21 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  nσ(m1 + 1) = nσ(m1) + 1, nσ(m2 + 1) = nσ(m2) + 1, nσ ≠ [], nσ(m2) ≠ nσ(m1), m1 < m2, ⌜m2⌝ + ⌜1⌝ < # nσ, nσ(m1) < nσ(m2), stutf(nσ), ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ(n + 1) = nσ(n) ∨ nσ(n + 1) = nσ(n) + 1, ∀ m0, nσ, n. m0 < m2.+1 → ⌜m0⌝ < # nσ ∧ n < nσ(m0) ∧ stutf(nσ) → (∃ m1. m1 < m0 ∧ nσ(m1) = n ∧ nσ(m1 + 1) = nσ(m1) + 1) ⊦ ∃ m0. ¬ m2 < m0 ∧ nσ(m0) = nσ(m1) ∧ nσ(m0 + 1) = nσ(m0) + 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 + 1 ≤ n ↔ m < n; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat

Proof step 22 from proof stutf-alltakenmax (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s) Used terms: m1