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Proof step 8 from proof restn-empty (previous step)

n = m + 1, restn(m, x0) = [], n ≤ # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ n = 0, n = 1 + # x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
restn(m, x0) = [], m < # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ m = # x0, x0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + 1 ≠ 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)