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Proof step 6 from proof unique-put (previous step)

a ≠ a0, ∀ a, d. unique_al(ax0[a, d]) ↔ unique_al(ax0) ⊦ unique_al((a0 × d0)' + ax0[a, d]) ↔ unique_al((a0 × d0)' + ax0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a, d. unique_al(ax0[a, d]) ↔ unique_al(ax0) ⊦ a = a0, ¬ a0 ∈ ax0 ∧ unique_al(ax0[a, d]) ↔ unique_al((a0 × d0)' + ax0)

Rule argument

Additional information

Used simplifier rules: cons: ⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax);%in-put: ⊦ a ∈ ax[b, d] ↔ a = b ∨ a ∈ ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;