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Proof step 6 from proof cons-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
disj(x, y0), ∀ x, a. disj(x, y0) → (disj(y0, a ' + x) ↔ ¬ a ∈ y0) ⊦ a = a0, a0 ∈ x, disj(y0, x), a ∈ y0

Rule argument

Additional information

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