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

r ∈ H1, H0 ⊥ H1 ⊦ r ∈ H0 ↔ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
r ∈ H1, r ∈ H0, H0 ⊥ H1 ⊦

Rule argument

Additional information