Proof step 1 from proof iterate-lemma (conclusion) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 2:

  ∀ n. ∃ nf. nf(0) = a ∧ (∀ m. m < n → f(nf(m)) = nf(m + 1)) ⊦ ∃ nf. nf(0) = a ∧ (∀ n. nf(n + 1) = f(nf(n)))

• Proof step 29:

  ⊦ ∀ n. ∃ nf. nf(0) = a ∧ (∀ m. m < n → f(nf(m)) = nf(m + 1)), ∃ nf. nf(0) = a ∧ (∀ n. nf(n + 1) = f(nf(n)))

Rule argument

Additional information

Used formulas: ∀ n. ∃ nf. nf(0) = a ∧ (∀ m. m < n → f(nf(m)) = nf(m + 1))

Proof step 2 from proof iterate-lemma (previous step) (back to tree)

User applied rule choice left

And obtained the following new goals

• Proof step 3:

  ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦ ∃ nf. nf(0) = a ∧ (∀ n. nf(n + 1) = f(nf(n)))

Rule argument

Additional information

Proof step 3 from proof iterate-lemma (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 4:

  ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦ nfgen(0)(0) = a ∧ (∀ n. nfgen(n + 1)(n + 1) = f(nfgen(n)(n)))

Rule argument

Additional information

Used terms: λ n. nfgen(n)(n)

Proof step 4 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦ nfgen(0)(0) = a ∧ nfgen(n + 1)(n + 1) = f(nfgen(n)(n))

Rule argument

Additional information

Proof step 5 from proof iterate-lemma (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 6:

  nfgen(0)(0) ≠ a, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

• Proof step 7:

  nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 6 from proof iterate-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: 0

Proof step 7 from proof iterate-lemma (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 8:

  nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m) ⊦

• Proof step 13:

  nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦ ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m)

Rule argument

Additional information

Used formulas: ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m)

Proof step 8 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 9:

  nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), n ≤ n ∧ n ≤ n + 1 → nfgen(n)(n) = nfgen(n + 1)(n), ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Used terms: n, n, n + 1

Proof step 9 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  nfgen(n)(n) = nfgen(n + 1)(n), nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m) ⊦

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; (s) ⊦ m ≤ m; (s)

Proof step 10 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 11:

  nfgen(n)(n) = nfgen(n + 1)(n), nfgen(n + 1)(0) = a, nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m. m < n + 1 → f(nfgen(n + 1)(m)) = nfgen(n + 1)(m + 1) ⊦

Rule argument

Additional information

Used terms: n + 1

Proof step 11 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  nfgen(n)(n) = nfgen(n + 1)(n), nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), ∀ n0. nfgen(n0)(0) = nfgen(n + 1)(0) ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m, n1, n2. m ≤ n1 ∧ m ≤ n2 → nfgen(n1)(m) = nfgen(n2)(m), ∀ m. ¬ n < m → f(nfgen(n + 1)(m)) = nfgen(n + 1)(m + 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 + 1 ↔ ¬ n < m; (s)

Proof step 12 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

Proof step 13 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  nfgen(n1)(m) ≠ nfgen(n2)(m), nfgen(n + 1)(n + 1) ≠ f(nfgen(n)(n)), m ≤ n1, m ≤ n2, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 14 from proof iterate-lemma (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 15:

  nfgen(n1)(m) ≠ nfgen(n2)(m), m ≤ n2, m ≤ n1, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 15 from proof iterate-lemma (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 16:

  nfgen(n1)(0) ≠ nfgen(n2)(0), 0 ≤ n1, 0 ≤ n2, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

• Proof step 20:

  nfgen(n1)(m0 +1) ≠ nfgen(n2)(m0 +1), m0 +1 ≤ n1, m0 +1 ≤ n2, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ nfgen, n1, n2, a, f. ¬(nfgen(n1)(m0) ≠ nfgen(n2)(m0) ∧ m0 ≤ n2 ∧ m0 ≤ n1 ∧ (∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)))) ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 16 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  nfgen(n1)(0) ≠ nfgen(n2)(0), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ m; (s)

Proof step 17 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 18:

  nfgen(n1)(0) = a, nfgen(n1)(0) ≠ nfgen(n2)(0), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m. m < n1 → f(nfgen(n1)(m)) = nfgen(n1)(m + 1) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 18 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  nfgen(n1)(0) ≠ nfgen(n2)(0), ∀ m. m < n1 → f(nfgen(n1)(m)) = nfgen(n1)(m + 1), ∀ n0. nfgen(n0)(0) = nfgen(n1)(0) ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 19 from proof iterate-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: n2

Proof step 20 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n1, m0 < n2, ∀ nfgen0, n, n0, a0, f0. nfgen0(n)(m0) = nfgen0(n0)(m0) ∨ ¬ m0 ≤ n0 ∨ ¬ m0 ≤ n ∨ ¬(∀ n0. nfgen0(n0)(0) = a0 ∧ (∀ m. m < n0 → f0(nfgen0(n0)(m)) = nfgen0(n0)(m + 1))), ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 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 + 1 ≤ n ↔ m < n; (s) ⊦ m +1 = m + 1; (s)

Proof step 21 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 22:

  nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n2, m0 < n1, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), nfgen(n1)(m0) = nfgen(n2)(m0) ∨ ¬ m0 ≤ n2 ∨ ¬ m0 ≤ n1 ∨ ¬(∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1))) ⊦

Rule argument

Additional information

Used terms: nfgen, n1, n2, a, f

Proof step 22 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n1, m0 < n2, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)

Proof step 23 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 24:

  nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n1)(0) = a, nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n2, m0 < n1, ∀ n0. nfgen(n0)(0) = a ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)), ∀ m. m < n1 → f(nfgen(n1)(m)) = nfgen(n1)(m + 1) ⊦

Rule argument

Additional information

Used terms: n1

Proof step 24 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n1, m0 < n2, ∀ m. m < n1 → f(nfgen(n1)(m)) = nfgen(n1)(m + 1), ∀ n0. nfgen(n0)(0) = nfgen(n1)(0) ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 25 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 26:

  nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n2, m0 < n1, m0 < n1 → f(nfgen(n1)(m0)) = nfgen(n1)(m0 + 1), ∀ n0. nfgen(n0)(0) = nfgen(n1)(0) ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Used terms: m0

Proof step 26 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  f(nfgen(n2)(m0)) = nfgen(n1)(m0 + 1), nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n1, m0 < n2, ∀ n0. nfgen(n0)(0) = nfgen(n1)(0) ∧ (∀ m. m < n0 → f(nfgen(n0)(m)) = nfgen(n0)(m + 1)) ⊦

Rule argument

Additional information

Proof step 27 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 28:

  f(nfgen(n2)(m0)) = nfgen(n1)(m0 + 1), nfgen(n1)(m0) = nfgen(n2)(m0), nfgen(n2)(0) = nfgen(n1)(0), nfgen(n1)(m0 + 1) ≠ nfgen(n2)(m0 + 1), m0 < n2, m0 < n1, ∀ m. m < n2 → f(nfgen(n2)(m)) = nfgen(n2)(m + 1) ⊦

Rule argument

Additional information

Used terms: n2

Proof step 28 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m0

Proof step 29 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ⊦ ∃ nf. nf(0) = a ∧ (∀ n. nf(n + 1) = f(nf(n))), ∃ nf. nf(0) = a ∧ (∀ m. m < n → f(nf(m)) = nf(m + 1))

Rule argument

Additional information

Proof step 30 from proof iterate-lemma (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 31:

  ⊦ ∃ nf. nf(0) = a ∧ (∀ m. m < n → f(nf(m)) = nf(m + 1))

Rule argument

Additional information

Proof step 31 from proof iterate-lemma (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 32:

  ⊦ ∃ nf. nf(0) = a ∧ (∀ m. m < 0 → f(nf(m)) = nf(m + 1))

• Proof step 34:

  ∀ a, f. ∃ nf. nf(0) = a ∧ (∀ m. m < n0 → f(nf(m)) = nf(m + 1)) ⊦ ∃ nf. nf(0) = a ∧ (∀ m. m < n0 +1 → f(nf(m)) = nf(m + 1))

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 32 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  ⊦ ∃ nf. nf(0) = a

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 33 from proof iterate-lemma (previous step) (back to tree)

User applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: λ n. a

Proof step 34 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 35:

  ∀ a, f. ∃ nf. nf(0) = a ∧ (∀ m. m < n0 → f(nf(m)) = nf(m + 1)) ⊦ ∃ nf. nf(0) = a ∧ (∀ m. ¬ n0 < m → f(nf(m)) = nf(m + 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 + 1 ↔ ¬ n < m; (s) ⊦ m +1 = m + 1; (s)

Proof step 35 from proof iterate-lemma (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 36:

  ∃ nf. nf(0) = a ∧ (∀ m. m < n0 → f(nf(m)) = nf(m + 1)) ⊦ ∃ nf. nf(0) = a ∧ (∀ m. ¬ n0 < m → f(nf(m)) = nf(m + 1))

Rule argument

Additional information

Used terms: a, f

Proof step 36 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦ ∃ nf0. nf0(0) = nf(0) ∧ (∀ m. ¬ n0 < m → f(nf0(m)) = nf0(m + 1))

Rule argument

Additional information

Proof step 37 from proof iterate-lemma (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 38:

  ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦ (0 = n0 + 1 ⊃ f(nf(n0));nf(0)) = nf(0) ∧ (∀ m. ¬ n0 < m → f((m = n0 + 1 ⊃ f(nf(n0));nf(m))) = (m + 1 = n0 + 1 ⊃ f(nf(n0));nf(m + 1)))

Rule argument

Additional information

Used terms: λ m. (m = n0 + 1 ⊃ f(nf(n0));nf(m))

Proof step 38 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  f((m = n0 + 1 ⊃ f(nf(n0));nf(m))) ≠ (m = n0 ⊃ f(nf(n0));nf(m + 1)), m ≤ n0, ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦

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 = m + n0 ↔ n = n0; (s) ⊦ m + 1 ≠ 0; (s)

Proof step 39 from proof iterate-lemma (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 40:

  f(nf(m)) ≠ (m = n0 ⊃ f(nf(n0));nf(m + 1)), m ≠ n0 + 1, m ≤ n0, ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦

Rule argument

Additional information

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

Proof step 40 from proof iterate-lemma (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 41:

  f(nf(m)) ≠ nf(m + 1), m ≠ n0, m ≠ n0 + 1, m ≤ n0, ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦

Rule argument

Additional information

Proof step 41 from proof iterate-lemma (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  f(nf(m)) ≠ nf(m + 1), m ≠ n0 + 1, m ≠ n0, m < n0, ∀ m. m < n0 → f(nf(m)) = nf(m + 1) ⊦

Rule argument

Additional information

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

Proof step 42 from proof iterate-lemma (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: m