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Proof step 2 from proof rev-in (previous step)

∀ a. a ∈ rev(x0) ↔ a ∈ x0 ⊦ a ∈ rev(a0 + x0) ↔ a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ rev(x0) ↔ a ∈ x0 ⊦ a ∈ rev(x0) ↔ a ∈ x0, a = a0

Rule argument

Additional information

Used simplifier rules: rev-r: ⊦ rev(a ' + x) = rev(x) + a;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%Last: ⊦ x + a = x + a '; from specification list%c: ⊦ a + x = a ' + x; from specification list