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Proof step 1 from proof empty (conclusion)

⊦ ¬ a ∈ []

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)

Rule argument

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

Additional information

Used simplifier rules: in-empty: ⊦ ¬ p ∈ []; from specification list[assoclista] Used terms: a, []