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Proof step 6 from proof lastoccurrence-key (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 7:
a ∈ ax0, ∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax1) ⊦ ∃ ax, d0, ax1. (a0 × d)' + ax0 = ax + (a × d0)' + ax1 ∧ ¬ a ∈ ax1, a ≠ a0 ∧ ¬ a ∈ ax0
Proof step 11:
∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax1) ⊦ a ∈ ax0, ∃ ax, d0, ax1. (a0 × d)' + ax0 = ax + (a × d0)' + ax1 ∧ ¬ a ∈ ax1, a ≠ a0 ∧ ¬ a ∈ ax0

Rule argument

fma: a ∈ ax0

Additional information

Used formulas: a ∈ ax0