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Proof step 6 from proof diff-consnotin2 (previous step)

¬ ¬ a0 ∈ a ' + x0, ∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ (a0 ' + x1) \ (a ' + x0) = (a0 ' + x1) \ x0, a = a0, a ∈ x1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ∈ x0, ∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ a = a0, x1 \ (a ' + x0) = x1 \ x0, a ∈ x1

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list