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Proof step 14 from proof firstoccurrence-key (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ [];% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ [] + ax = ax;% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1; Used terms: [], d, ax + (a0 × d0)' + ax1