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Proof step 4 from proof at-put (previous step)

∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ a = b, (p + ax0)[b, d][a] = (p + ax0)[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ a = b, (p ' + ax0)[b, d][a] = (p ' + ax0)[a]

Rule argument

Additional information

Used simplifier rules: ⊦ p + ax = p ' + ax;