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Proof step 1 from proof inone (conclusion)

⊦ a ∈ (b × d0)' ↔ a = b

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)
Proof step 3:
a ∈ (b × d0)' ↔ (∃ d. a × d ∈ (b × d0)') ⊦ a ∈ (b × d0)' ↔ a = b

Rule argument

spec: assoclist inst: name: inassoc seq: ⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax) subst: [a, ax] → [a, (b × d0)'] flags: current, []

Additional information

Used terms: a, (b × d0)'