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Proof step 68 from proof infinite-append-butlast (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 12 terms: m

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m