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Proof step 15 from proof lastn-app (previous step)

n0 + # y = m + 1, ∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ y = [], lastn(m, (x + y).butlast) + (x + y).last ' = lastn(n0, x) + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n0 + # y = m + 1, ∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ y = [], lastn(m, x + y.butlast) + y.last ' = lastn(n0, x) + y

Rule argument

Additional information

Used simplifier rules: lapp: ⊦ y ≠ [] → (x + y).last = y.last; from specification list-last%a: ⊦ (x + y) + z = x + y + z; from specification list%bapp: ⊦ y ≠ [] → (x + y).butlast = x + y.butlast; from specification list-last