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Proof step 3 from proof sw-alloc-put (previous step)

r1 ∈ H ⊦ (H ++ r0)[r0, o0][r1, o1] = (H[r1, o1] ++ r0)[r0, o0], r0 = null, r0 ∈ 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:
r1 ∈ H ⊦ r0 = null, r0 ∈ H, ∀ r. (r ∈ (H ++ r0)[r0, o0][r1, o1] ↔ r ∈ (H[r1, o1] ++ r0)[r0, o0]) ∧ (r ∈ (H ++ r0)[r0, o0][r1, o1] → (H ++ r0)[r0, o0][r1, o1][r] = (H[r1, o1] ++ r0)[r0, o0][r])

Rule argument

spec: <toplevel> name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [(H[r1, o1] ++ r0)[r0, o0], (H ++ r0)[r0, o0][r1, o1]] flags: <all paths>

Additional information

Used terms: (H[r1, o1] ++ r0)[r0, o0], (H ++ r0)[r0, o0][r1, o1]