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Proof step 11 from proof sym (previous step)

a ∈ y, ∀ y. disj(x0, y) → disj(y, x0), ∀ a0. ¬(a0 ∈ a ' + x0 ∧ a0 ∈ y) ⊦ disj(y, a ' + x0), disj(x0, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ y, ∀ y. disj(x0, y) → disj(y, x0), ∀ a0. a0 ≠ a ∧ ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ disj(x0, y), disj(y, a ' + x0)

Rule argument

Additional information

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