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Proof step 5 from proof in-diffs (previous step)

r ∈ (H -- r0) \ rs0 ↔ r ∈ H -- r0 ∧ ¬ r ∈ rs0, ∀ r, H. r ∈ H \ rs0 ↔ r ∈ H ∧ ¬ r ∈ rs0 ⊦ r ∈ (H -- r0) \ rs0 ↔ r ∈ H ∧ r ≠ r0 ∧ ¬ r ∈ rs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

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