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Proof step 3 from proof abs-keys-orphans2 (previous step)

abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ⊦ orphans(fs0, ri0) = orphans(fs, ri)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0)
Proof step 5:
abs-keys(orphans(fs0, ri0)) = abs-keys(orphans(fs, ri)) ⊦ ∀ key. key ∈ orphans(fs0, ri0) ↔ key ∈ orphans(fs, ri)

Rule argument

spec: set-basic, inst: node-index name: Extension seq: ⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0) subst: [ks0, ks] → [orphans(fs, ri), orphans(fs0, ri0)] flags: <all paths>

Additional information

Used terms: orphans(fs, ri), orphans(fs0, ri0)