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Proof step 6 from proof in-assoclist-01 (
previous step
)
a ∈ ax0, ∀ a. a ∈ ax0 → a × ax0[a] ∈ ax0 ⊦ a = a0, a × ax0[a] = a0 × d, a × ax0[a] ∈ ax0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a
Additional information
Used terms: a