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Proof step 5 from proof insnotin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a, d. ax0[a, d] = ax0 + (a × d)' ∨ a ∈ ax0 ⊦ a = a0, ax0[a, d] = ax0 + (a × d)', a ∈ 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];%app-08: ⊦ p ' + ax = p ' + ax0 ↔ ax = ax0; from specification list[assoclista]%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]