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Proof step 6 from proof in-diffs (
previous step
)
r ∈ (H -- r0) \ rs0 ↔ r ≠ r0 ∧ r ∈ H ∧ ¬ r ∈ rs0, ∀ r, H. r ∈ H \ rs0 ↔ r ∈ H ∧ ¬ r ∈ rs0 ⊦ r ∈ (H -- r0) \ rs0 ↔ r ∈ H ∧ r ≠ r0 ∧ ¬ r ∈ rs0
User applied rule case distinction
And closed the goal!
Rule argument
pos: right 1
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