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Proof step 5 from proof disj-distr-uni (previous step)

H1 ⊥ H ⊦ (∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∪ H) ↔ H0 ⊥ H1 ∧ H0 ⊥ H

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
H1 ⊥ H, ∀ r. ¬ r ∈ H0 ∨ ¬ r ∈ H1 ∧ ¬ r ∈ H ⊦ H0 ⊥ H1 ∧ H0 ⊥ H

Rule argument

Additional information

Used simplifier rules: in-uni: ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1);%disj-l: ⊦ r ∈ H0 ∧ H0 ⊥ H1 → (r ∈ H1 ↔ false);