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Proof step 2 from proof lf-cmp-peb (previous step)

apebs[n0] = apeb0, n0 < n + m, cmp(apebs, apeb, n, m), n ≤ n0 ⊦ apeb0 = apeb

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n0 < n + m, cmp(apebs, apeb, n, m), n ≤ n0 ⊦ apebs[n0] = apeb, m = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic