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Proof step 43 from proof nextl-lemma-butlast-nonempty (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 44:
lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ lf(m + 1) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last), m = n0

Rule argument

pos: right 1

Additional information

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