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

⊦ ∀ r0. (r0 ∈ (H -- r) \ rs ↔ r0 ∈ (H \ rs) -- r) ∧ (r0 ∈ (H -- r) \ rs → ((H -- r) \ rs)[r0] = ((H \ rs) -- r)[r0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r0 ∈ H ⊦ ((H -- r) \ rs)[r0] = H[r0], r0 = r, r0 ∈ rs

Rule argument

Additional information

Used simplifier rules: in-diffs: ⊦ r ∈ H \ rs ↔ r ∈ H ∧ ¬ r ∈ rs;%in-del: ⊦ r ∈ H -- r0 ↔ r ≠ r0 ∧ r ∈ H;%at-del: r ∈ H ⊦ r ≠ r0 → (H -- r0)[r] = H[r];%at-diffs: ⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r];