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Proof step 4 from proof diff-consnotin2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, x0. x1 \ (a ' + x0) = x1 \ x0 ∨ a ∈ x1 ⊦ (a0 ' + x1) \ (a ' + x0) = (a0 ' + x1) \ x0, a = a0, a ∈ x1

Rule argument

Additional information

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