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Proof step 6 from proof union-eq-empty (previous step)

∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ ¬ r ∈ zh0 ∧ zh1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ zh = zh0 ↔ (∀ r. (r ∈ zh ↔ r ∈ zh0) ∧ (r ∈ zh → zh[r] = zh0[r]))
Proof step 8:
∅ = zh0 ∪ zh1, zh0 ⊥ zh1 ⊦ ¬ r ∈ zh0 ∧ (∀ r. (r ∈ zh1 ↔ r ∈ ∅) ∧ (r ∈ zh1 → zh1[r] = (∅)[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] flags: <all paths>

Additional information

Used terms: ∅, zh1