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Proof step 5 from proof insfirst (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ a = a0, (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0, a ∈ ax1

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];%app-08: ⊦ p ' + ax = p ' + ax0 ↔ ax = ax0; from specification list[assoclista]%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]