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Proof step 18 from proof lastn-app (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n0 + # y = m + 1, lastn(n0 + (# y - 1), x + y.butlast) = lastn(n0, x) + y.butlast, ∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ lastn(m, x + y.butlast) + y.last ' = lastn(n0, x) + y, y = []

Rule argument

Additional information

Used simplifier rules: ⊦ x ≠ [] → # x.butlast = # x - 1; (s)