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Proof step 7 from proof diff-onenonmember (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ a. x0 \ a ' = x0 ∨ a ∈ x0 ⊦ x0 \ a ' = x0, a = a0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s)% ⊦ a + x = a ' + x; (s)