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Proof step 14 from proof inapp (
previous step
)
(∃ d. a × d ∈ ax) ∨ (∃ d. a × d ∈ ay) ⊦ ∃ d. a × d ∈ ax ∨ a × d ∈ ay
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 15
:
a × d ∈ ax ∨ a × d0 ∈ ay ⊦ ∃ d. a × d ∈ ax ∨ a × d ∈ ay
Rule argument
Additional information