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Proof step 7 from proof restn-empty (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
restn(m, x0) = [], m < # x0, ∀ n0. n0 < # x0 → restn(n0, x0) = [] → n0 = # x0 ⊦ m = # x0, x0 = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2