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Proof step 1 from proof cons (conclusion)

⊦ unique_al((a × d)' + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ p + ax = p ' + ax
Proof step 3:
⊦ unique_al((a × d) + ax) ↔ ¬ a ∈ ax ∧ unique_al(ax)

Rule argument

spec: list inst: assoclista name: c seq: ⊦ p + ax = p ' + ax subst: [ax, p] → [ax, a × d] flags: rewrite

Additional information

Used terms: ax, a × d