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Proof step 3 from proof firstoc (previous step)

∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2), a ∈ p + ax0 ⊦ ∃ ax2, d, ax1. p + ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2), a ∈ p ' ∨ a ∈ ax0 ⊦ ∃ ax, d, ax1. p ' + ax0 = ax + (a × d)' + ax1 ∧ ¬ a ∈ ax

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]