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Proof step 5 from proof in (previous step)

b = a0, a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ a = b, a = a0, a ∈ (a0 ' + x0) -l b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ x0, ∀ a, b. a ∈ x0 → a = b ∨ a ∈ x0 -l b ⊦ a = a0, a ∈ x0 -l a0

Rule argument

Additional information

Used simplifier rules: del-y: ⊦ (a ' + x) -l a = x -l a;%a: ⊦ (x + y) + z = x + y + z; from specification list