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Proof step 3 from proof cons-02 (previous step)

∀ x, a. disj(x, y0) → (disj(y0, a ' + x) ↔ ¬ a ∈ y0), disj(x, a0 + y0) ⊦ disj(a0 + y0, a ' + x) ↔ ¬ a ∈ a0 + y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
disj(x, y0), ∀ x, a. disj(x, y0) → (disj(y0, a ' + x) ↔ ¬ a ∈ y0) ⊦ disj(a0 ' + y0, a ' + x) ↔ a ≠ a0 ∧ ¬ a ∈ y0, a0 ∈ x

Rule argument

Additional information

Used simplifier rules: app-02: ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z);%one-01: ⊦ disj(x, a ') ↔ ¬ a ∈ x;%a: ⊦ (x + y) + z = x + y + z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list