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Proof step 7 from proof cons-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
disj(x, y), a ∈ y, ∀ a0. ¬ a0 ∈ x ∧ a0 ≠ a ∨ ¬ a0 ∈ y ⊦

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)