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Proof step 8 from proof unique (previous step)

unique_al(ax0 + (a × d)') ⊦ a0 = a, ax0 = [], a0 ∈ ax0, ¬ ax0.first.1 ∈ ax0.rest ∧ unique_al(ax0.rest), ∃ a, d. unique_al(ax0 + (a × d)')

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: a, d

Additional information

Used terms: a, d