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Proof step 13 from proof firstoccurrence-key (previous step)

⊦ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ a0 ∈ ax, ∃ ax3, d1, ax2. (a0 × d)' + ax + (a0 × d0)' + ax1 = ax3 + (a0 × d1)' + ax2 ∧ ¬ a0 ∈ ax3

Rule argument

Additional information

Used simplifier rules: in-append: ⊦ a ∈ ax + ay ↔ a ∈ ax ∨ a ∈ ay;%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;