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Proof step 11 from proof lookup-02 (previous step)

a0 × d0 ∈ ax0, unique_al(ax0), ∀ a0, d0. a0 × d0 ∈ ax0 → ax0[a0] = d0 ⊦ a0 = a, ax0[a0] = d0, a0 × d0 = a × d, a ∈ ax0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a0, d0

Additional information

Used terms: a0, d0