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Proof step 11 from proof sw-del-add-new (previous step)

r0 ∈ H ⊦ ((H ++ r)[r, o] \ rs)[r0] = H[r0], r = null, r ∈ H, r ∈ rs, r0 ∈ rs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ r ∈ H ∧ ¬ r ∈ rs → (H \ rs)[r] = H[r]
Proof step 13:
r0 ∈ H, r0 ∈ (H ++ r)[r, o] ⊦ (H ++ r)[r, o][r0] = H[r0], r = null, r ∈ rs, r ∈ H, r0 ∈ rs

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)[r, o]] flags: <all paths>

Additional information

Used simplifier rules: in-write: ⊦ r ∈ H[r0, o] ↔ r ∈ H;%in-alloc: r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; Used terms: r0, rs, (H ++ r)[r, o]