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Proof step 7 from proof ex (previous step)

∃ ax0, ax1. ax = ax0 + ((a × d) + ax1) ⊦ ∃ d, ay, az. ax = ay + (a × d)' + az

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ ∃ d0, ay, az. ax1 + (a × d)' + ax0 = ay + (a × d0)' + az

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%c: ⊦ p + ax = p ' + ax; from specification list[assoclista]