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Proof step 5 from proof unique-02 (previous step)

unique_al((a0 × d1)' + ax0), ∀ a, d, d0. unique_al(ax0) ∧ a × d ∈ ax0 ∧ a × d0 ∈ ax0 → d = d0, a × d = a0 × d1 ∨ a × d ∈ ax0, a × d0 = a0 × d1 ∨ a × d0 ∈ ax0 ⊦ d = d0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
unique_al(ax0), a × d = a0 × d1 ∨ a × d ∈ ax0, a × d0 = a0 × d1 ∨ a × d0 ∈ ax0, ∀ a1, d2, d3. a1 × d2 ∈ ax0 ∧ a1 × d3 ∈ ax0 → d2 = d3 ⊦ d = d0, a0 ∈ ax0

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]