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

dups(rev(x0)) ↔ dups(x0) ⊦ dups(rev(a + x0)) ↔ dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
dups(rev(x0)) ↔ dups(x0) ⊦ dups(rev(x0) + a ') ↔ a ∈ x0 ∨ dups(x0)

Rule argument

Additional information

Used simplifier rules: rev-r: ⊦ rev(a ' + x) = rev(x) + a;%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%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