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

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

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, r0 = r, r ∈ H, ∀ r1. (r1 ∈ ((H -- r0) ++ r)[r, o] ↔ r1 ∈ (H ++ r)[r, o] -- r0) ∧ (r1 ∈ ((H -- r0) ++ r)[r, o] → ((H -- r0) ++ r)[r, o][r1] = ((H ++ r)[r, o] -- r0)[r1])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [(H ++ r)[r, o] -- r0, ((H -- r0) ++ r)[r, o]] flags: <all paths>

Additional information

Used terms: (H ++ r)[r, o] -- r0, ((H -- r0) ++ r)[r, o]