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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%in-rec: ⊦ p ∈ p0 ' + ax ↔ ¬ ¬(p = p0 ∨ p ∈ ax); from specification list[assoclista]%c: ⊦ p + ax = p ' + ax; from specification list[assoclista]