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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2