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Proof step 3 from proof in-assoclist (previous step)

∀ a. a ∈ ax0 → a × ax0[a] ∈ ax0, a ∈ p + ax0 ⊦ a × (p + ax0)[a] ∈ p + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. a ∈ ax0 → a × ax0[a] ∈ ax0 ⊦ a × (p ' + ax0)[a] = p, a × (p ' + ax0)[a] ∈ ax0, ¬ a ∈ p ' ∧ ¬ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);% ⊦ ¬ p ∈ p0 ' + ax ↔ p ≠ p0 ∧ ¬ p ∈ ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p + ax = p ' + ax;