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Proof step 6 from proof unique (previous step)

unique_al((p + ax0) + (a × d)'), ∀ a, d. unique_al(ax0 + (a × d)') → ax0 = [] ∨ ¬ ax0.head.1 ∈ ax0.tail ∧ unique_al(ax0.tail) ⊦ p + ax0 = [], ¬ (p + ax0).head.1 ∈ (p + ax0).tail ∧ unique_al((p + ax0).tail)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
unique_al((p ' + ax0) + (a × d)'), (∃ a, d. unique_al(ax0 + (a × d)')) → ax0 = [] ∨ ¬ ax0.head.1 ∈ ax0.tail ∧ unique_al(ax0.tail) ⊦ ¬ p.1 ∈ ax0 ∧ unique_al(ax0)

Rule argument

Additional information

Used simplifier rules: ⊦ (p ' + ax).tail = ax; (s)% ⊦ (p ' + ax).head = p; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclist]% ⊦ p ' + ax ≠ []; (s)% ⊦ p + ax = p ' + ax; (s)