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

r0 ∈ H ⊦ ((H -- r) \ rs)[r0] = H[r0], r0 = r, r0 ∈ rs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r]

Rule argument

spec: <toplevel> name: at-diffs seq: ⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r] subst: [r, rs, H] → [r0, rs, H -- r] flags: <all paths>

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: r0, rs, H -- r