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Proof step 64 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%lez: ⊦ 0 ≤ m; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%leprf: ⊦ m ≤ m - 1 → m = 0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ls: n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat