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Proof step 5 from proof inapp (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)
Proof step 7:
⊦ (∃ d. a × d ∈ ax ∨ a × d ∈ ay) ↔ (∃ d. a × d ∈ ax) ∨ a ∈ ay

Rule argument

spec: assoclist inst: name: inassoc seq: ⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax) subst: [ax, a] → [ax, a] flags: rewrite, current

Additional information

Used terms: ax, a