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Proof step 2 from proof disj-l (previous step)

r ∈ H0, H0 ⊥ H1 ⊦ r ∈ H1 ↔ false

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: disj-r: ⊦ r ∈ H1 ∧ H0 ⊥ H1 → (r ∈ H0 ↔ false);