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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

Proof step 6:
∀ x. # lastn(n0, x) = n0 ⊦ # lastn(n0, x.butlast) = n0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ x + a = x + a '; (s)