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Proof step 3 from proof disj-distr-uni (previous step)

H1 ⊥ H ⊦ H0 ⊥ H1 ∪ H ↔ H0 ⊥ H1 ∧ H0 ⊥ H

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1)
Proof step 5:
H1 ⊥ H ⊦ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∪ H) ↔ H0 ⊥ H1 ∧ H0 ⊥ H

Rule argument

spec: <toplevel> name: disj seq: ⊦ H0 ⊥ H1 ↔ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1) subst: [H1, H0] → [H1 ∪ H, H0] flags: <all paths>

Additional information

Used terms: H1 ∪ H, H0