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Proof step 11 from proof firstoc (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]