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Proof step 5 from proof putset-same (previous step)

∀ x. n0 < # x → x[n0, x[n0]] = x ∨ x = [], n0 +1 < # x ⊦ x[n0 +1, x[n0 +1]] = x, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 + 1 < # x, ∀ x. n0 < # x → x[n0, x[n0]] = x ∨ x = [] ⊦ x = [], x[n0 + 1, x.rest[n0]] = x

Rule argument

Additional information

Used simplifier rules: get-rec-01: ⊦ x ≠ [] → x[n + 1] = x.rest[n];%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2