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Proof step 20 from proof replayorphans-prepost (previous step)

ri0 ⊖ ks0 = ri1 ⊖ ro1, ks0.inodes?, ro1.inodes?, key ∈ ro1, key ∈ ks0, ks0 ⊆ ro1 ⊦ # ro1 = 0, # ks0 = 0, ks0 = ∅, ro1 = ∅, ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key) = ri1 ⊖ ro1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ ri = ri0 ↔ (∀ key. (key ∈ ri ↔ key ∈ ri0) ∧ ri[key] = ri0[key])
Proof step 22:
ri0 ⊖ ks0 = ri1 ⊖ ro1, ro1.inodes?, ks0.inodes?, key ∈ ks0, key ∈ ro1, ks0 ⊆ ro1 ⊦ # ks0 = 0, # ro1 = 0, ks0 = ∅, ro1 = ∅, ∀ key0. (key0 ∈ ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key) ↔ key0 ∈ ri1 ⊖ ro1) ∧ (((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key))[key0] = (ri1 ⊖ ro1)[key0]

Rule argument

spec: store, inst: node-index name: Extension seq: ⊦ ri = ri0 ↔ (∀ key. (key ∈ ri ↔ key ∈ ri0) ∧ ri[key] = ri0[key]) subst: [ri0, ri] → [ri1 ⊖ ro1, ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key)] flags: <all paths>

Additional information

Used terms: ri1 ⊖ ro1, ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key)