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Proof step 44 from proof nextl-lemma (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 45:
m ≠ n0 + 1, lf(0) = nextl(0, a), m < n0, ∀ m. m < n0 → lf(m + 1) = nextl(m + 1, lf(m).last) ⊦ lf(m + 1) = nextl(m + 1, lf(m).last), m = n0, n0 = 0

Rule argument

pos: right 1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2