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Proof step 3 from proof notin-dels (previous step)

adr ∈ ri ⊖ ks ⊦ adr ∈ ri

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr)
Proof step 5:
∃ key. key ∈ ri ⊖ ks ∧ (ri ⊖ ks)[key] = adr ⊦ adr ∈ ri

Rule argument

spec: store+set, inst: node-index name: Data-in seq: ⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr) subst: [ri, adr] → [ri ⊖ ks, adr] flags: <all paths>

Additional information

Used terms: ri ⊖ ks, adr