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

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 4

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