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Proof step 5 from proof firstoc (previous step)

⊦ (∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2)) → ¬ a0 ∈ (a × d)' ∧ ¬ a0 ∈ ax0 ∨ (∃ ax, d0, ax1. ¬ a0 ∈ ax ∧ (a × d)' + ax0 = ax + (a0 × d0)' + ax1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2) ⊦ ∃ ax, d0, ax1. ¬ a0 ∈ ax ∧ (a × d)' + ax0 = ax + (a0 × d0)' + ax1, ¬ a0 ∈ ax0 ∧ a0 ≠ a

Rule argument

Additional information

Used simplifier rules: inone: ⊦ a ∈ (b × d0)' ↔ a = b;