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Proof step 7 from proof app-02 (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
disj(x, z), disj(x, y0), ∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, y0 + z) ↔ disj(x, z), a ∈ x
Proof step 10:
disj(x, y0), ∀ x, z. disj(x, y0 + z) ↔ disj(x, y0) ∧ disj(x, z) ⊦ disj(x, z), disj(x, y0 + z) ↔ disj(x, z), a ∈ x

Rule argument

fma: disj(x, z)

Additional information

Used formulas: disj(x, z)