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Proof step 5 from proof unique (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
unique_al((p ' + ax0) + (a × d)'), (∃ a, d. unique_al(ax0 + (a × d)')) → ax0 = [] ∨ ¬ ax0.first.1 ∈ ax0.rest ∧ unique_al(ax0.rest) ⊦ ¬ p.1 ∈ ax0 ∧ unique_al(ax0)

Rule argument

Additional information

Used simplifier rules: c-07: ⊦ (p ' + ax).rest = ax; from specification list[assoclista]%c-08: ⊦ (p ' + ax).first = p; from specification list[assoclista]%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]%e-01: ⊦ p ' + ax ≠ []; from specification list[assoclista]%c: ⊦ p + ax = p ' + ax; from specification list[assoclista]