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Proof step 2 from proof last-ex (previous step)

a ∈ x ⊦ a = x.last, ∃ y, z. x = y + a ' + z ∧ z ≠ []

User applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '
Proof step 4:
x = y0 + a0 ' ∧ x ≠ [], a ∈ x ⊦ a = a0, ∃ y, z. x = y + a ' + z ∧ z ≠ []

Rule argument

spec: list-last inst: name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x, a0, y0] flags: rewrite, current

Additional information

Used simplifier rules: in-empty: ⊦ ¬ a ∈ []; from specification list Used terms: x, a0, y0