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Proof step 3 from proof sw-del-add-new (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 5:
⊦ r = null, r ∈ rs, r ∈ H, ∀ r0. (r0 ∈ (H ++ r)[r, o] \ rs ↔ r0 ∈ ((H \ rs) ++ r)[r, o]) ∧ (r0 ∈ (H ++ r)[r, o] \ rs → ((H ++ r)[r, o] \ rs)[r0] = ((H \ rs) ++ r)[r, o][r0])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [((H \ rs) ++ r)[r, o], (H ++ r)[r, o] \ rs] flags: <all paths>

Additional information

Used terms: ((H \ rs) ++ r)[r, o], (H ++ r)[r, o] \ rs