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Proof step 6 from proof firstoc (previous step)

∀ 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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
a = a0, ∀ 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
Proof step 9:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2) ⊦ a = a0, ∃ ax, d0, ax1. ¬ a0 ∈ ax ∧ (a × d)' + ax0 = ax + (a0 × d0)' + ax1, ¬ a0 ∈ ax0 ∧ a0 ≠ a

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0