Proof step 1 from proof infinite-append-lem (conclusion) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 2:

  ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Rule argument

Additional information

Proof step 2 from proof infinite-append-lem (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < 0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

• Proof step 5:

  ∀ lf. ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m))) ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 +1 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Rule argument

Additional information

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

Proof step 3 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  ⊦ ∃ ii. ii(0) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < 0; (s)

Proof step 4 from proof infinite-append-lem (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: λ n. n

Proof step 5 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  ∀ lf. ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m))) ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. ¬ n0 < m → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(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 < n + 1 ↔ ¬ n < m; (s) ⊦ m +1 = m + 1; (s)

Proof step 6 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 7:

  ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m))) ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. ¬ n0 < m → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Rule argument

Additional information

Used terms: lf

Proof step 7 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  ii(0) = 0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. ¬ n0 < m → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Rule argument

Additional information

Proof step 8 from proof infinite-append-lem (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 9:

  ii(0) = 0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ (0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(0)) = 0 ∧ (∀ m. ¬ n0 < m → (m + 1 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m + 1)) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ( (λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))))

Rule argument

Additional information

Used terms: λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1)), λ n1. (n1 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(n1))

Proof step 9 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  ii(0) = 0, m ≤ n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ (m = n0 ⊃ ii(n0) + # lf(n0);ii(m + 1)) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)))

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 10 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 11:

  ii(0) = 0, n0 = m, m ≤ n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ (m = n0 ⊃ ii(n0) + # lf(n0);ii(m + 1)) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)))

• Proof step 25:

  ii(0) = 0, n0 ≠ m, m ≤ n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ (m = n0 ⊃ ii(n0) + # lf(n0);ii(m + 1)) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)))

Rule argument

Additional information

Used formulas: n0 = m

Proof step 11 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  ii(0) = 0, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦ ((λ n1. (n1 < ii(m) + # lf(m) ∧ ¬ n1 < ii(m) ⊃ lf(m)[n1 - ii(m)];nf(n1))) from ii(m)) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)) ≤ ii(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 ↔ n = 0; (s) ⊦ m + n = m + n0 ↔ n = n0; (s) ⊦ m ≤ m; (s)

Proof step 12 from proof infinite-append-lem (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 13:

  ii(0) = 0, ((λ n1. (n1 < ii(m) + # lf(m) ∧ ¬ n1 < ii(m) ⊃ lf(m)[n1 - ii(m)];nf(n1))) from ii(m)) to # lf(m) ≠ lf(m), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

• Proof step 15:

  ii(0) = 0, m0 ≤ m, ¬ (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)) ≤ ii(m), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

Proof step 13 from proof infinite-append-lem (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

Rule argument

Additional information

Used simplifier rules: ⊦ #(nf to n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ ((nf from n))(m) = nf(m + n); (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ m + n < m; (s) ⊦ m + n - m = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: lf(m), ( (λ n0. (n0 < ii(m) + # lf(m) ∧ ¬ n0 < ii(m) ⊃ lf(m)[n0 - ii(m)];nf(n0))) from ii(m)) to # lf(m)

Proof step 14 from proof infinite-append-lem (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  ii(0) = 0, ii(m) < (m0 = m + 1 ⊃ ii(m) + # lf(m);ii(m0)), m0 ≤ m, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

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

Proof step 16 from proof infinite-append-lem (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 17:

  ii(0) = 0, m0 ≠ m + 1, ii(m) < ii(m0), m0 ≤ m, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ 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 17 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  ii(0) = 0, m0 ≠ m + 1, ii(m) < ii(m0), m0 ≤ m, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

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

Proof step 18 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 19:

  ii(0) = 0, m0 ≠ m, m0 ≠ m + 1, ii(m) < ii(m0), m0 ≤ m, ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m ≤ m; (s) ⊦ ¬ n < n; (s) Used formulas: m0 = m

Proof step 19 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  ii(0) = 0, m0 ≠ m + 1, m0 ≠ m, m0 < m, ii(m) < ii(m0), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

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 20 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 21:

  ii(0) = 0, m0 ≠ m + 1, m0 ≠ m, ii(m) < ii(m0), m0 < m, m - 1 < m → ii((m - 1) + 1) = ii(m - 1) + # lf(m - 1) ∧ (nf from ii(m - 1)) to # lf(m - 1) = lf(m - 1) ∧ (∀ m1. m1 ≤ m - 1 → ii(m1) ≤ ii(m - 1)), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

Used terms: m - 1

Proof step 21 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  ii(0) = 0, ii((m - 1) + 1) = ii(m - 1) + # lf(m - 1), (nf from ii(m - 1)) to # lf(m - 1) = lf(m - 1), m0 ≠ m + 1, m0 ≠ m, m0 < m, ii(m) < ii(m0), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m1. m1 < m → ii(m1) ≤ ii(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 + 1 ≤ n ↔ m < n; (s) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) ⊦ m < n → n ≠ 0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)

Proof step 22 from proof infinite-append-lem (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 23:

  ii(m2 + 1) = ii(m2) + # lf(m2), ii(0) = 0, m = m2 + 1, (nf from ii(m2)) to # lf(m2) = lf(m2), m ≠ 0, m0 ≠ m + 1, m0 ≠ m, ii(m) < ii(m0), m0 < m, ∀ m1. m1 < m → ii(m1) ≤ ii(m2), ∀ m0. m0 < m → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ n < 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m2

Proof step 23 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  ii(m2 + 1) = ii(m2) + # lf(m2), ii(0) = 0, (nf from ii(m2)) to # lf(m2) = lf(m2), m0 ≠ 2 + m2, m0 ≠ m2 + 1, ii(m2 + 1) < ii(m0), m0 ≤ m2, ∀ m1. ¬ m2 < m1 → ii(m1) ≤ ii(m2), ∀ m. ¬ m2 < m → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m1. m1 ≤ m → ii(m1) ≤ ii(m)) ⊦

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (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 < n ↔ ¬ n ≤ m; (ss) ⊦ m + 1 ≠ 0; (s)

Proof step 24 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (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 Used terms: m0

Proof step 25 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 26:

  ii(0) = 0, n0 ≠ m, m < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ ii(m + 1) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)))

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 26 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 27:

  ii(0) = 0, m ≠ n0 + 1, n0 ≠ m, m < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ ii(m + 1) = (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m))) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ (m = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m)))

Rule argument

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (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 Used formulas: m = n0 + 1

Proof step 27 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  ii(0) = 0, m ≠ n0 + 1, n0 ≠ m, m < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ ii(m + 1) = ii(m) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ ii(m))

Rule argument

Additional information

Proof step 28 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 29:

  ii(0) = 0, m ≠ n0 + 1, n0 ≠ m, m < n0, m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦ ii(m + 1) = ii(m) + # lf(m) ∧ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ ii(m))

Rule argument

Additional information

Used terms: m

Proof step 29 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, m < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦ ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m) = lf(m) ∧ (m0 ≤ m → (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ ii(m))

Rule argument

Additional information

Proof step 30 from proof infinite-append-lem (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 31:

  ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, ((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m) ≠ lf(m), m < n0, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

• Proof step 60:

  ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, m < n0, m0 ≤ m, ¬ (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)) ≤ ii(m), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Proof step 31 from proof infinite-append-lem (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

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

• Proof step 33:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, m < n0, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), #(((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m)) ≠ # lf(m) ∨ (∃ n. ¬ ( n < #(((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m)) → (((λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))) from ii(m)) to # lf(m))[n] = lf(m)[n])) ⊦

Rule argument

Additional information

Used terms: lf(m), ( (λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)] ;nf(n1) )) from ii(m)) to # lf(m)

Proof step 32 from proof infinite-append-lem (previous step) (back to tree)

A spec-lemma

Proof step 33 from proof infinite-append-lem (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ nf from n = (λ m. nf(n + m))

• Proof step 35:

  ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, m < n0, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), #((λ m0. (ii(m) + m0 < ii(n0) + # lf(n0) ∧ ¬ ii(m) + m0 < ii(n0) ⊃ lf(n0)[ii(m) + m0 - ii(n0)];nf(ii(m) + m0))) to # lf(m)) ≠ # lf(m) ∨ (∃ n. ¬ ( n < # ((λ m0. (ii(m) + m0 < ii(n0) + # lf(n0) ∧ ¬ ii(m) + m0 < ii(n0) ⊃ lf(n0)[ii(m) + m0 - ii(n0)];nf(ii(m) + m0))) to # lf(m)) → ((λ m0. (ii(m) + m0 < ii(n0) + # lf(n0) ∧ ¬ ii(m) + m0 < ii(n0) ⊃ lf(n0)[ii(m) + m0 - ii(n0)];nf(ii(m) + m0))) to # lf(m))[n] = lf(m)[n])) ⊦

Rule argument

Additional information

Used terms: ii(m), λ n1. (n1 < ii(n0) + # lf(n0) ∧ ¬ n1 < ii(n0) ⊃ lf(n0)[n1 - ii(n0)];nf(n1))

Proof step 34 from proof infinite-append-lem (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), (ii(m) + n < ii(n0) + # lf(n0) ∧ ¬ ii(m) + n < ii(n0) ⊃ lf(n0)[ii(m) + n - ii(n0)];nf(ii(m) + n)) ≠ lf(m)[n], lf(m) ≠ [], m ≠ n0 + 1, n0 ≠ m, n < # lf(m), m < n0, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ #(nf to n) = n; (s)

Proof step 36 from proof infinite-append-lem (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 37:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[ii(m) + n - ii(n0)] ≠ lf(m)[n], lf(m) ≠ [], m ≠ n0 + 1, n0 ≠ m, m < n0, n < # lf(m), ii(m) + n < ii(n0) + # lf(n0), ¬ ii(m) + n < ii(n0), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

• Proof step 56:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), lf(m) ≠ [], m ≠ n0 + 1, n0 ≠ m, nf(ii(m) + n) ≠ lf(m)[n], m < n0, n < # lf(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ¬ ii(m) + n < ii(n0) + # lf(n0) ∨ ii(m) + n < ii(n0) ⊦

Rule argument

Additional information

Proof step 37 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[ii(m) + n - ii(n0)] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ m, m < n0, n < # lf(m), ii(m) + n < ii(n0) + # lf(n0), ii(n0) ≤ ii(m) + n, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(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 ⊦ # x = 0 ↔ x = []; (s) ⊦ m < m + n ↔ n ≠ 0; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 38 from proof infinite-append-lem (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 39:

  ii(m) + n = n1 + ii(n0), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ m, ii(m) + n < ii(n0) + # lf(n0), n < # lf(m), m < n0, ii(n0) ≤ ii(m) + n, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: ii(m) + n, ii(n0), n1

Proof step 39 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  ii(m) + n = n1 + ii(n0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ m, n1 < # lf(n0), m < n0, n < # lf(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (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 ≤ m + n; (s) ⊦ m + n < m + n0 ↔ n < n0; (s)

Proof step 40 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 41:

  ii(m) + n = n1 + ii(n0), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ m, n < # lf(m), m < n0, n1 < # lf(n0), n0 - 1 < n0 → ii((n0 - 1) + 1) = ii(n0 - 1) + # lf(n0 - 1) ∧ (nf from ii(n0 - 1)) to # lf(n0 - 1) = lf(n0 - 1) ∧ (∀ m0. m0 ≤ n0 - 1 → ii(m0) ≤ ii(n0 - 1)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Used terms: n0 - 1

Proof step 41 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  ii(m) + n = n1 + ii(n0), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii((n0 - 1) + 1) = ii(n0 - 1) + # lf(n0 - 1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(n0 - 1)) to # lf(n0 - 1) = lf(n0 - 1), lf(n0)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ m, n1 < # lf(n0), m < n0, n < # lf(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. m0 < n0 → ii(m0) ≤ ii(n0 - 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) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) ⊦ m < n → n ≠ 0; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ ¬ n < n; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)

Proof step 42 from proof infinite-append-lem (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 43:

  ii(m) + n = n1 + ii(n0), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, n0 = m1 + 1, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), lf(n0)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(n0) ≠ [], m ≠ n0 + 1, n0 ≠ 0, n0 ≠ m, n < # lf(m), m < n0, n1 < # lf(n0), ∀ m0. m0 < n0 → ii(m0) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ n < 0; (s) n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m1

Proof step 43 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 44:

  ii(m) + n = n1 + ii(m1 + 1), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m1 + 1) ≠ [], lf(m) ≠ [], m ≠ 2 + m1, n1 < # lf(m1 + 1), n < # lf(m), m ≤ m1, ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (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 < n ↔ ¬ n ≤ m; (ss) ⊦ m + 1 ≠ 0; (s)

Proof step 44 from proof infinite-append-lem (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 45:

  ii(m) + n = n1 + ii(m1 + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m1 + 1) ≠ [], lf(m) ≠ [], m ≠ 2 + m1, n < # lf(m), n1 < # lf(m1 + 1), m ≤ m1, ¬ m1 < m + 1 → ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

Rule argument

Additional information

Used terms: m + 1

Proof step 45 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 46:

  ii(m) + n = n1 + ii(m1 + 1), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m1 + 1) ≠ [], lf(m) ≠ [], m ≠ 2 + m1, n1 < # lf(m1 + 1), n < # lf(m), m ≤ m1, m < m1 → ii(m + 1) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(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 < n + 1 ↔ ¬ n < m; (s)

Proof step 46 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 47:

  ii(m) + n = n1 + ii(m1 + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), m1 = m, (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m1 + 1) ≠ [], lf(m) ≠ [], m ≠ 2 + m1, n < # lf(m), n1 < # lf(m1 + 1), m ≤ m1, m < m1 → ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

• Proof step 49:

  ii(m) + n = n1 + ii(m1 + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m1 + 1) ≠ [], lf(m) ≠ [], m ≠ 2 + m1, m1 ≠ m, n < # lf(m), n1 < # lf(m1 + 1), m ≤ m1, m < m1 → ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

Rule argument

Additional information

Used formulas: m1 = m

Proof step 47 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 48:

  ii(m) + n = n1 + ii(m + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), lf(m + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m + 1) ≠ [], n < # lf(m), n1 < # lf(m + 1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m < m0 → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

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

Proof step 48 from proof infinite-append-lem (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = m + n0 ↔ n = n0; (s) lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic ⊦ m + m0 < n → ¬ n + n0 < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 49 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  ii(m) + n = n1 + ii(m1 + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, m < m1, n < # lf(m), n1 < # lf(m1 + 1), ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

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 50 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 51:

  ii(m) + n = n1 + ii(m1 + 1), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, n1 < # lf(m1 + 1), n < # lf(m), m < m1, ii(m) + n < ii(m1), ii(m + 1) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

• Proof step 53:

  ii(m) + n = n1 + ii(m1 + 1), ii(m1 + 1) = ii(m1) + # lf(m1), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m1)) to # lf(m1) = lf(m1), (nf from ii(m)) to # lf(m) = lf(m), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, n1 < # lf(m1 + 1), n < # lf(m), m < m1, ii(m + 1) ≤ ii(m1), ¬ ii(m) + n < ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

Used formulas: ii(m) + n < ii(m1)

Proof step 51 from proof infinite-append-lem (previous step) (back to tree)

User applied rule subst equation

And obtained the following new goals

• Proof step 52:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, n1 + ii(m1 + 1) < ii(m1), m < m1, n < # lf(m), n1 < # lf(m1 + 1), ii(m + 1) ≤ ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

Rule argument

Additional information

Proof step 52 from proof infinite-append-lem (previous step) (back to tree)

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

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

Proof step 53 from proof infinite-append-lem (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 54:

  ii(m) + n = n1 + ii(m1 + 1), ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), ii(m1 + 1) = ii(m1) + # lf(m1), (nf from ii(m)) to # lf(m) = lf(m), (nf from ii(m1)) to # lf(m1) = lf(m1), m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, m < m1, n < # lf(m), n1 < # lf(m1 + 1), ii(m) + n < ii(m + 1), ii(m + 1) ≤ ii(m1), ¬ ii(m) + n < ii(m1), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m0. ¬ m1 < m0 → ii(m0 + 1) = ii(m0) + # lf(m0) ∧ (nf from ii(m0)) to # lf(m0) = lf(m0) ∧ (∀ m1. m1 ≤ m0 → ii(m1) ≤ ii(m0)), ∀ m0. ¬ m1 < m0 → ii(m0) ≤ ii(m1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used formulas: ii(m) + n < ii(m + 1)

Proof step 54 from proof infinite-append-lem (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 55:

  m1 + 1 ≠ m, lf(m1 + 1)[n1] ≠ lf(m)[n], lf(m) ≠ [], lf(m1 + 1) ≠ [], m ≠ 2 + m1, m1 ≠ m, ii(m) + n < ii(m + 1), ii(m + 1) ≤ ii(m1), ¬ ii(m) + n < ii(m1) ⊦

Rule argument

Additional information

Proof step 55 from proof infinite-append-lem (previous step) (back to tree)

User applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m < n → n ≠ 0; (s) let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ m < n ↔ ¬ n ≤ m; (ss)

Proof step 56 from proof infinite-append-lem (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 57:

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

• Proof step 58:

  #((nf from ii(m)) to # lf(m)) = # lf(m), ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, lf(m) ≠ [], m ≠ n0 + 1, n0 ≠ m, nf(ii(m) + n) ≠ lf(m)[n], n < # lf(m), m < n0, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ n. n < #((nf from ii(m)) to # lf(m)) → ((nf from ii(m)) to # lf(m))[n] = lf(m)[n], ¬ ii(m) + n < ii(n0) + # lf(n0) ∨ ii(m) + n < ii(n0) ⊦

Rule argument

Additional information

Used terms: lf(m), (nf from ii(m)) to # lf(m)

Proof step 57 from proof infinite-append-lem (previous step) (back to tree)

A spec-lemma

Proof step 58 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 59:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), lf(m) ≠ [], m ≠ n0 + 1, n0 ≠ m, nf(ii(m) + n) ≠ lf(m)[n], m < n0, n < # lf(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ n1. n1 < # lf(m) → nf(n1 + ii(m)) = lf(m)[n1], ¬ ii(m) + n < ii(n0) + # lf(n0) ∨ ii(m) + n < ii(n0) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ((nf from n))(m) = nf(m + n); (s) m < n ⊦ (nf to n)[m] = nf(m); (s) ⊦ m < n → (n ≤ m ↔ false); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ #(nf to n) = n; (s)

Proof step 59 from proof infinite-append-lem (previous step) (back to tree)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic Used terms: n

Proof step 60 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 61:

  ii(m + 1) = ii(m) + # lf(m), ii(0) = 0, (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, n0 ≠ m, m0 < n0, ii(m) < (m0 = n0 + 1 ⊃ ii(n0) + # lf(n0);ii(m0)), m < n0, m0 ≤ m, ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m), ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss)

Proof step 61 from proof infinite-append-lem (previous step) (back to tree)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

• Proof step 62:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, m0 ≠ n0 + 1, n0 ≠ m, m < n0, m0 < n0, ii(m) < ii(m0), m0 ≤ m, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

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

Proof step 62 from proof infinite-append-lem (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 63:

  ii(0) = 0, ii(m + 1) = ii(m) + # lf(m), (nf from ii(m)) to # lf(m) = lf(m), m ≠ n0 + 1, m0 ≠ n0 + 1, n0 ≠ m, m < n0, m0 < n0, ii(m) < ii(m0), m0 ≤ m, ∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)), ∀ m0. m0 ≤ m → ii(m0) ≤ ii(m) ⊦

Rule argument

Additional information

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

Proof step 63 from proof infinite-append-lem (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