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Proof step 7 from proof lf-cmp-peb (previous step)

n0 < n + m, ∀ n0. n0 < m → apebs[n + n0] = apeb, ∃ n1. n + n1 = n0 ⊦ apebs[n0] = apeb, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n1 < m, ∀ n0. n0 < m → apebs[n + n0] = apeb ⊦ apebs[n + n1] = apeb

Rule argument

Additional information

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