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Proof step 3 from proof insfirst (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a, d, ax0, d0. (ax1 + (a × d)' + ax0)[a, d0] = ax1 + (a × d0)' + ax0 ∨ a ∈ ax1 ⊦ ((p ' + ax1) + (a × d)' + ax0)[a, d0] = (p ' + ax1) + (a × d0)' + ax0, a ∈ p ', a ∈ ax1

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%c: ⊦ p + ax = p ' + ax; from specification list[assoclista]