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Proof step 27 from proof del-uni (previous step)

H0 ⊥ H1, H0 -- r ⊥ H1 -- r ⊦ r0 ≠ r ∧ (r0 ∈ H0 ∨ r0 ∈ H1) ↔ r0 ∈ H0 -- r ∨ r0 ∈ H1 -- r

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
H0 -- r ⊥ H1 -- r, H0 ⊥ H1, r0 ≠ r ∧ r0 ∈ H0 ∨ r0 ≠ r ∧ r0 ∈ H1 ⊦ r0 ≠ r ∧ (r0 ∈ H0 ∨ r0 ∈ H1)

Rule argument

Additional information

Used simplifier rules: in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;