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Proof step 3 from proof lastn-neq-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last
Proof step 5:
lastn(n0, x.butlast) + x.last = b ' + y, a ∈ y, n0 < # x, ∀ x, a, y, b. n0 ≤ # x ∧ a ∈ y ∧ ¬ a ∈ x → lastn(n0, x) ≠ b ' + y ⊦ x = [], a ∈ x

Rule argument

spec: inst: name: lastN-rec seq: ⊦ lastn(n + 1, x) = lastn(n, x.butlast) + x.last subst: [x, n] → [x, n0] flags: rewrite, current

Additional information

Used terms: x, n0