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Proof step 7 from proof uniquelen (previous step)

unique_al(p ' + ax), ∀ a. ¬ a ∈ p ' ∧ ¬ a ∈ ax ⊦ p.1 ∈ ax, p.1 ∈ p '

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ a = p.1 ∧ d = p.2 ↔ p = a × d subst: [p, a, d] → [p, a0, d] rotate: T spec: pair inst: name: elim

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ a = p.1 ∧ d = p.2 ↔ p = a × d; (elim) Used terms: p, a0, d