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Proof step 5 from proof insfirst (previous step)

⊦ (∀ a, d, ax1, d0. (ax0 + (a × d)' + ax1)[a, d0] = ax0 + (a × d0)' + ax1 ∨ a ∈ ax0) → (((a × d)' + ax0) + (a0 × d1)' + ax1)[a0, d0] = ((a × d)' + ax0) + (a0 × d0)' + ax1 ∨ a0 ∈ ax0 ∨ a0 ∈ (a × d)'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;%put-no: ⊦ a ≠ b → ((b × d0)' + ax)[a, d] = (b × d0)' + ax[a, d];% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;