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

∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, (a ' + y0) + z) ↔ disj(x, a ' + y0) ∧ disj(x, z)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ a ∈ x, disj(x, (a ' + y0) + z) ↔ disj(x, a ' + y0) ∧ disj(x, z)

Rule argument

fma: a ∈ x

Additional information

Used simplifier rules: in: ⊦ disj(x, y) ∧ a ∈ x → ¬ a ∈ y;%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: a ∈ x