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Proof step 11 from proof at-put (previous step)

a ≠ a0, ∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ b = a0, ax0[b, d][a] = ((a0 × d0)' + ax0)[a], a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ a, b, d. a = b ∨ ax0[b, d][a] = ax0[a] ⊦ a = b, a = a0, b = a0, ax0[b, d][a] = ax0[a]

Rule argument

Additional information

Used simplifier rules: assoc-no: ⊦ a ≠ b → ((b × d)' + ax)[a] = ax[a];% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;