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Proof step 7 from proof unique (previous step)

⊦ unique_al(((a × d)' + ax0) + (a0 × d0)') ∧ ((∃ a, d. unique_al(ax0 + (a × d)')) → unique_al(ax0)) → unique_al((a × d)' + ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
unique_al(ax0 + (a0 × d0)') ⊦ a = a0, a ∈ ax0, unique_al(ax0), ∃ a, d. unique_al(ax0 + (a × d)')

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;