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Proof step 3 from proof eq-union-disj (previous step)

zh0 ∪ zh = zh1 ∪ zh, zh1 ⊥ zh, zh0 ⊥ zh ⊦ zh0 = zh1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ zh = zh0 ↔ (∀ r. (r ∈ zh ↔ r ∈ zh0) ∧ (r ∈ zh → zh[r] = zh0[r]))
Proof step 5:
zh0 ∪ zh = zh1 ∪ zh, zh0 ⊥ zh, zh1 ⊥ zh ⊦ ∀ r. (r ∈ zh0 ↔ r ∈ zh1) ∧ (r ∈ zh0 → zh0[r] = zh1[r])

Rule argument

spec: heap, inst: zheap name: ext seq: ⊦ zh = zh0 ↔ (∀ r. (r ∈ zh ↔ r ∈ zh0) ∧ (r ∈ zh → zh[r] = zh0[r])) subst: [zh0, zh] → [zh1, zh0] flags: <all paths>

Additional information

Used terms: zh1, zh0