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Proof step 5 from proof insnotin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, d. ax0[a, d] = ax0 + (a × d)' ∨ a ∈ ax0 ⊦ a0 = a, ax0[a0, d0] = ax0 + (a0 × d0)', a0 ∈ ax0

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;