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Proof step 49 from proof nextl-lemma-butlast-nonempty (previous step)

(m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) = [], lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 50:
lf(0) = a ', lf(m) = [], m ≤ n0, ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ⊦ m = n0 + 1

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic