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Proof step 2 from proof in-01 (
previous step
)
a × d ∈ ax ⊦ a ∈ ax
User applied rule insert lemma
And obtained the following new goals
Proof step 3
:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)
Proof step 4
:
a ∈ ax ↔ (∃ d. a × d ∈ ax), a × d ∈ ax ⊦ a ∈ ax
Rule argument
spec: <toplevel> name: inassoc seq: ⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax) subst: [a, ax] → [a, ax]
Additional information
Used terms: a, ax