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Proof step 9 from proof lastn-app (previous step)

∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ lastn((n0 + # y)-1, (x + y).butlast) + (x + y).last = lastn(n0, x) + y, n0 + # y = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ n0 + # y = 0, lastn(n0 + # y - 1, (x + y).butlast) + (x + y).last ' = lastn(n0, x) + y

Rule argument

Additional information

Used simplifier rules: ⊦ x + a = x + a '; (s)% ⊦ n -1 = n - 1; (s)