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Proof step 4 from proof inapp (previous step)

⊦ (∃ d. a × d ∈ ax + ay) ↔ a ∈ ax ∨ a ∈ ay

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (∃ d. a × d ∈ ax ∨ a × d ∈ ay) ↔ a ∈ ax ∨ a ∈ ay

Rule argument

Additional information

Used simplifier rules: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);