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Proof step 3 from proof replay-nonorphans-trunc (previous step)

⊦ (ri ⊖ ro) \ datakeys(key, n, ri) = (ri \ datakeys(key, n, ri)) ⊖ ro, key ∈ ro

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:
⊦ key ∈ ro, ∀ key0. (key0 ∈ (ri ⊖ ro) \ datakeys(key, n, ri) ↔ key0 ∈ (ri \ datakeys(key, n, ri)) ⊖ ro) ∧ (key0 ∈ (ri ⊖ ro) \ datakeys(key, n, ri) → ((ri ⊖ ro) \ datakeys(key, n, ri))[key0] = ((ri \ datakeys(key, n, ri)) ⊖ ro)[key0])

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 \ datakeys(key, n, ri)) ⊖ ro, (ri ⊖ ro) \ datakeys(key, n, ri)] flags: <all paths>

Additional information

Used terms: (ri \ datakeys(key, n, ri)) ⊖ ro, (ri ⊖ ro) \ datakeys(key, n, ri)