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Proof step 4 from proof lastn-neq-01 (previous step)

lastn(n0 + 1, x) = b ' + y, n0 < # x, a ∈ y, ∀ x, b, y, a. lastn(n0, x) = b ' + y ∧ a ∈ y ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last
Proof step 6:
lastn(n0, x.butlast) + x.last = b ' + y, n0 < # x, a ∈ y, ∀ x, b, y, a. lastn(n0, x) = b ' + y ∧ a ∈ y ∧ n0 ≤ # x → a ∈ x ⊦ x = [], a ∈ x

Rule argument

spec: <toplevel> name: lastN-rec seq: ⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last subst: [x, n] → [x, n0] flags: <all paths>

Additional information

Used terms: x, n0