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Proof step 4 from proof at-diffs (previous step)

r ∈ H, ∀ r, H. r ∈ H → (H \ rs0)[r] = H[r] ∨ r ∈ rs0 ⊦ (H \ (rs0 ++ r0))[r] = H[r], r ∈ rs0 ++ r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
r ∈ H, ∀ r, H. r ∈ H → (H \ rs0)[r] = H[r] ∨ r ∈ rs0 ⊦ ((H -- r0) \ rs0)[r] = H[r], 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); (ss)