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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ a + x = a ' + x; from specification list