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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ x. # lastn(n0, x) = n0 ⊦ # lastn(n0 + 1, x) = n0 + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)