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Proof step 11 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 ⊦ lastn(m, (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 12:
n0 + # y = m + 1, ∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ lastn(m, (x + y).butlast) + (x + y).last ' = lastn(n0, x) + y

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 + 1 ≠ 0; (s)