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Proof step 61 from proof infinite-append-butlast (previous step)

# x = m0 + 1, a ' + y = lf(m + 1).head ', ((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, ii(m) = m0 + 1, ii(n) = m0 + 1, (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), a ∈ a ' + y, n ≤ m, ii(n0 + 1) ≤ m0 + 1, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).tail) ∧ ¬ m0 + 1 < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).tail) = (n1 = 0 ⊃ lf(0);lf(n1).tail), ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).head, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ a ' + y = [], m = 0, y = [] ∧ m0 = (n = 0 ⊃ 0;m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
# x = m0 + 1, ((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, ii(m) = m0 + 1, ii(n) = m0 + 1, (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), n ≤ m, ii(n0 + 1) ≤ m0 + 1, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).tail) ∧ ¬ m0 + 1 < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).tail) = (n1 = 0 ⊃ lf(0);lf(n1).tail), ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).head, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ m = 0, m0 = (n = 0 ⊃ 0;m0)

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' = b ' + x ↔ a = b ∧ x = []; (ws)