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Proof step 3 from proof del-del-again (previous step)

inodekey(m) ∈ ks ⊦ # ks = 0, ks = ∅, (ri ⊖ ks) ⊖ inodekey(m) = ri ⊖ ks

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:
inodekey(m) ∈ ks ⊦ # ks = 0, ks = ∅, ∀ key. (key ∈ (ri ⊖ ks) ⊖ inodekey(m) ↔ key ∈ ri ⊖ ks) ∧ (key ∈ (ri ⊖ ks) ⊖ inodekey(m) → ((ri ⊖ ks) ⊖ inodekey(m))[key] = (ri ⊖ ks)[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 ⊖ ks, (ri ⊖ ks) ⊖ inodekey(m)] flags: <all paths>

Additional information

Used terms: ri ⊖ ks, (ri ⊖ ks) ⊖ inodekey(m)