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Proof step 3 from proof lastn-length (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last
Proof step 5:
∀ x. # lastn(n0, x) = n0 ⊦ #(lastn(n0, x.butlast) + x.last) = n0 + 1

Rule argument

spec: list-del inst: name: lastN-rec seq: ⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last subst: [x, n] → [x, n0] flags: rewrite, current

Additional information

Used terms: x, n0