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Proof step 59 from proof infinite-append-butlast (previous step)

# x = m0 + 1, ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), a0 = lf(m + 1).head, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), ii(0) = 0, ii(m) = m0 + 1, ii(n) = m0 + 1, lf(m) = a ' + y, lf(m) = a0 + y0, (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), y0 = [], a ∈ lf(m), 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) ⊦ lf(m) = [], m = 0, y = [] ∧ m0 = (n = 0 ⊃ 0;m0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
# x = m0 + 1, ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(n) = m0 + 1, ii(m) = m0 + 1, ii(0) = 0, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), lf(m) = a ' + y, lf(m) = lf(m + 1).head ', (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).tail) = (n0 = 0 ⊃ lf(0);lf(n0).tail), a ∈ lf(m), (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, ii(n0 + 1) ≤ m0 + 1, n ≤ m, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).head, ∀ 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) ⊦ lf(m) = [], m = 0, y = [] ∧ m0 = (n = 0 ⊃ 0;m0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)% ⊦ a + x = a ' + x; (s)