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Proof step 2 from proof ex (previous step)

⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax)
Proof step 4:
⊦ (∃ d. a × d ∈ ax) ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)

Rule argument

spec: <toplevel> name: inassoc seq: ⊦ a ∈ ax ↔ (∃ d. a × d ∈ ax) subst: [ax, a] → [ax, a] flags: <all paths>

Additional information

Used terms: ax, a