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

⊦ ∀ r. (r ∈ (H \ rs) \ rs ↔ r ∈ H \ rs) ∧ (r ∈ (H \ rs) \ rs → ((H \ rs) \ rs)[r] = (H \ rs)[r])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r ∈ H ⊦ ((H \ rs) \ rs)[r] = H[r], r ∈ rs

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ H \ rs ↔ r ∈ H ∧ ¬ r ∈ rs;%at-diffs: ⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r];