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Proof step 4 from proof cons-02 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z)
Proof step 6:
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

Rule argument

spec: inst: name: app-02 seq: ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z) subst: [z, y, x] → [x, a ', a0 ' + y0] flags: rewrite, current

Additional information

Used terms: x, a ', a0 ' + y0