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Proof step 3 from proof in-diffs (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ 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: diffs-ins: ⊦ H0 \ (rs ++ r) = (H0 -- r) \ rs;% ⊦ r ∈ rs ++ r0 ↔ r = r0 ∨ r ∈ rs; (s)