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Proof step 10 from proof lookup-append (previous step)

a0 ∈ ax0, ∀ a, ay. a ∈ ax0 → (ax0 + ay)[a] = ax0[a] ⊦ a0 = a, (ax0 + ay)[a0] = ax0[a0]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0, ay

Additional information

Used terms: a0, ay