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Proof step 2 from proof in-last-rest (previous step)

⊦ a ∈ (x + b ').tail ↔ a = b ∨ a ∈ x.tail, x = []

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: rest-01: ⊦ x ≠ [] → (x + y).tail = x.tail + y;%a: ⊦ (x + y) + z = x + y + z;%in-append: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y;%in-17: ⊦ a ∈ b ' ↔ a = b;