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Proof step 9 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a ∈ ax0, ∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2) ⊦ a = a0, ∃ ax, d0, ax1. (a0 × d)' + ax0 = ax + (a × d0)' + ax1 ∧ ¬ a ∈ ax

Proof step 9 from proof firstoc (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information