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Proof step 5 from proof precise-mapex (previous step)

(∅ ++ r)[r, o] ∪ H0 = (∅ ++ r)[r, o0] ∪ H ⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H, r ∈ H0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 7:
∀ r0. (r0 ∈ (∅ ++ r)[r, o] ∪ H0 ↔ r0 ∈ (∅ ++ r)[r, o0] ∪ H) ∧ (r0 ∈ (∅ ++ r)[r, o] ∪ H0 → ((∅ ++ r)[r, o] ∪ H0)[r0] = ((∅ ++ r)[r, o0] ∪ H)[r0]) ⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H0, r ∈ H

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, o0] ∪ H, (∅ ++ r)[r, o] ∪ H0] flags: <all paths>

Additional information

Used terms: (∅ ++ r)[r, o0] ∪ H, (∅ ++ r)[r, o] ∪ H0