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Proof step 1 from proof unique-add (conclusion)

⊦ unique_al(ax + p ') ↔ ¬ p.1 ∈ ax ∧ unique_al(ax)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 2:
p = a × d ⊦ unique_al(ax + p ') ↔ ¬ a ∈ ax ∧ unique_al(ax)

Rule argument

seq: ⊦ a = p.1 ∧ d = p.2 ↔ p = a × d subst: [p, a, d] → [p, a, d] rotate: T spec: pair inst: name: elim

Additional information

Used simplifier rules: ⊦ a = p.1 ∧ d = p.2 ↔ p = a × d; (elim) Used terms: p, a, d