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Proof step 65 from proof infinite-append-butlast (previous step)

((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(0) = 0, (nf from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ (nf from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = (n = 0 ⊃ lf(0);lf(n).tail), ∀ n. nf(ii(n + 1) - 1) = lf(n + 1).head, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∃ m. ii(m) ≤ n2 + 1 ∧ n2 + 1 < ii(m + 1) ⊦ ∃ m0. (m0 = 0 ⊃ 0;ii(m0) - 1) ≤ n2 ∧ n2 < ii(m0 + 1) - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(0) = 0, (nf from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(m) < ii(m + 1), n2 + 1 < ii(m + 1), ii(m) ≤ n2 + 1, ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ (nf from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = (n = 0 ⊃ lf(0);lf(n).tail), ∀ n. nf(ii(n + 1) - 1) = lf(n + 1).head, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [] ⊦ ∃ m0. (m0 = 0 ⊃ 0;ii(m0) - 1) ≤ n2 ∧ n2 < ii(m0 + 1) - 1

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