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Proof step 11 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ a = a0, a0 ∈ ax, ∃ ax2, d1, ax3. ¬ a0 ∈ ax2 ∧ (a × d)' + ax + (a0 × d0)' + ax1 = ax2 + (a0 × d1)' + ax3

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ ax = ax0 + p ' + ax1 → p ∈ ax; (forward)