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Proof step 35 from proof index-update-adr-in (previous step)

# kx0 = # ax1, key0 ∈ ri ⊦ key0 = key, dups(kx0), ri[key0] ∈ ri[key, adr1] \ kx0.toSet, ri[key0] ∈ ax1, key ∈ kx0, ri[key0] ∈ ri[key, adr1][kx0, ax1], key0 ∈ kx0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 36:
⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr)
Proof step 37:
# kx0 = # ax1, key0 ∈ ri ⊦ key0 = key, dups(kx0), key0 ∈ kx0, ri[key0] ∈ ri[key, adr1][kx0, ax1], key ∈ kx0, ri[key0] ∈ ax1, ∃ key1. key1 ∈ ri[key, adr1] \ kx0.toSet ∧ (ri[key, adr1] \ kx0.toSet)[key1] = ri[key0]

Rule argument

spec: store+set, inst: node-index name: Data-in seq: ⊦ adr ∈ ri ↔ (∃ key. key ∈ ri ∧ ri[key] = adr) subst: [ri, adr] → [ri[key, adr1] \ kx0.toSet, ri[key0]] flags: <all paths>

Additional information

Used terms: ri[key, adr1] \ kx0.toSet, ri[key0]