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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: r, H -- r0

Additional information

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