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Proof step 6 from proof unique-01 (previous step)

unique_al(ax0 + (a × d)'), a ∈ ax0, unique_al(ax0), ∀ a1, d0. ¬ a1 ∈ ax0 ∨ ¬ unique_al(ax0 + (a1 × d0)') ⊦ a0 ∈ ax0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a, d

Additional information

Used terms: a, d