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Proof step 5 from proof lastn-length (previous step)

∀ x. # lastn(n0, x) = n0 ⊦ #(lastn(n0, x.butlast) + x.last) = n0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: 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%a: ⊦ (x + y) + z = x + y + z; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%Last: ⊦ x + a = x + a '; from specification list