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Proof step 11 from proof ex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ ∃ d0. a × d0 ∈ ay ∨ a × d0 = a × d ∨ a × d0 ∈ az

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]% ⊦ p ∈ p0 ' ↔ p = p0; (s)% ⊦ p ∈ ax + ax0 ↔ p ∈ ax ∨ p ∈ ax0; (s)