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Proof step 3 from proof unique-02 (previous step)

∀ a, d0, d. a × d0 ∈ ax0 ∧ a × d ∈ ax0 ∧ unique_al(ax0) → d = d0 ⊦ a × d0 ∈ p + ax0 ∧ a × d ∈ p + ax0 ∧ unique_al(p + ax0) → d = d0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ∈ p0 ' ↔ p = p0;% ⊦ p ∈ ax + ax0 ↔ ¬ ¬(p ∈ ax ∨ p ∈ ax0);% ⊦ p + ax = p ' + ax;