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Proof step 1 from proof firstn-prefix-04 (conclusion)

⊦ # x = (1 + # z) + n0 → (¬ firstn(1 + n0, x) ⊑ x ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# x = (1 + # z) + n0 ⊦ x.first ' + firstn(n0, x.rest) ⊑ x

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list