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Proof step 46 from proof infinite-append-butlast (previous step)

# x = ii(n), ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1 = ii(n0 + 1), ii(0) = 0, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).tail), (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), ii(n0 + 1) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n), ∀ 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) ⊦ 1 + (n = 0 ⊃ 0;ii(n) - 1) = ii(n)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m < n → n ≠ 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic