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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lastN-rec-01: ⊦ lastn(n + 1, x + a ') = lastn(n, x) + a ';%app-03: ⊦ x + y = z + y ↔ x = z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%Last: ⊦ x + a = x + a '; from specification list