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Proof step 28 from proof mapex-new (previous step)

H ++ r = (∅ ++ r)[r, o] ∪ H0, P(H0) ⊦ r = null, P(H), r ∈ H, r ∈ H0, ∀ r. (r ∈ H0 ↔ r ∈ H) ∧ (r ∈ H0 → H0[r] = H[r])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 30:
P(H0), ∀ r0. (r0 ∈ H ++ r ↔ r0 ∈ (∅ ++ r)[r, o] ∪ H0) ∧ (r0 ∈ H ++ r → (H ++ r)[r0] = ((∅ ++ r)[r, o] ∪ H0)[r0]) ⊦ r = null, P(H), r ∈ H0, r ∈ H, ∀ r. (r ∈ H0 ↔ r ∈ H) ∧ (r ∈ H0 → H0[r] = H[r])

Rule argument

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

Additional information

Used terms: (∅ ++ r)[r, o] ∪ H0, H ++ r