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Proof step 48 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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