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Proof step 54 from proof infinite-append-butlast (previous step)

ii(n) = m0 + 1, # x = m0 + 1, 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), lf(m).last = lf(m + 1).first, lf(m) = a ' + y, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, n ≤ m, ii(n0 + 1) ≤ m0 + 1, a ∈ lf(m), ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).rest) ∧ ¬ m0 + 1 < 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, n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = m0 + 1, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ lf(m) = [], y = [] ∧ (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: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat