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Proof step 2 from proof first-length (previous step)

∀ x. # firstn(n0, x) = n0 ⊦ # firstn(n0 +1, x) = n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: firstN-rec: ⊦ firstn(n + 1, x) = x.first + firstn(n, x.rest);%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%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%succ: ⊦ m +1 = m + 1; from specification nat-basic2