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Proof step 12 from proof firstoccurrence-key (previous step)

⊦ a ∈ ax, ∃ ax3, d1, ax2. (a0 × d)' + ax + (a × d0)' + ax1 = ax3 + (a × d1)' + ax2 ∧ ¬ a ∈ ax3

User applied rule exists right

And obtained the following new goals

Proof step 13:
⊦ a ∈ ax, (a0 × d)' + ax + (a × d0)' + ax1 = ((a0 × d)' + ax) + (a × d0)' + ax1 ∧ ¬ a ∈ (a0 × d)' + ax, ∃ ax3, d1, ax2. (a0 × d)' + ax + (a × d0)' + ax1 = ax3 + (a × d1)' + ax2 ∧ ¬ a ∈ ax3

Rule argument

pos: right 2 terms: (a0 × d)' + ax, d0, ax1 abort: F discard: F computed: T

Additional information

Used terms: (a0 × d)' + ax, d0, ax1