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Proof step 41 from proof nextl-lemma-butlast-nonempty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
lf(0) = a ', m ≤ n0, ∀ n, a. nextl(n, a) ≠ [], ∀ m. m < n0 → lf(m + 1) = nextl(m, lf(m).last) ∧ lf(m) ≠ [] ⊦ (m = n0 ⊃ nextl(n0, lf(n0).last);lf(m + 1)) = nextl(m, (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)).last) ∧ (m = n0 + 1 ⊃ nextl(n0, lf(n0).last);lf(m)) ≠ []

Rule argument

Additional information

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