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Proof step 2 from proof in-put (previous step)

∀ a, b, d. a ∈ ax0[b, d] ↔ a = b ∨ a ∈ ax0 ⊦ a ∈ (p + ax0)[b, d] ↔ a = b ∨ a ∈ p + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a, b, d. a ∈ ax0[b, d] ↔ a = b ∨ a ∈ ax0 ⊦ a ∈ (p ' + ax0)[b, d] ↔ a = b ∨ a ∈ p ' ∨ a ∈ ax0

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p + ax = p ' + ax;