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Proof step 3 from proof drop-le (previous step)

key1 < key0 ⊦ ri[key0, adr] |≤ key1 = ri |≤ key1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ri = ri0 ↔ (∀ key. (key ∈ ri ↔ key ∈ ri0) ∧ (key ∈ ri → ri[key] = ri0[key]))
Proof step 5:
key1 < key0 ⊦ ∀ key. (key ∈ ri[key0, adr] |≤ key1 ↔ key ∈ ri |≤ key1) ∧ (key ∈ ri[key0, adr] |≤ key1 → (ri[key0, adr] |≤ key1)[key] = (ri |≤ key1)[key])

Rule argument

spec: store, inst: node-index name: k-Extension seq: ⊦ ri = ri0 ↔ (∀ key. (key ∈ ri ↔ key ∈ ri0) ∧ (key ∈ ri → ri[key] = ri0[key])) subst: [ri0, ri] → [ri |≤ key1, ri[key0, adr] |≤ key1] flags: <all paths>

Additional information

Used terms: ri |≤ key1, ri[key0, adr] |≤ key1