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Proof step 26 from proof keys-same-fs (previous step)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ fs = fs0 ↔ (∀ adr. (adr ∈ fs ↔ adr ∈ fs0) ∧ fs[adr] = fs0[adr])
Proof step 28:
∀ adr0. (adr0 ∈ fs0 | (ax.toSet ++ adr) ↔ adr0 ∈ fs1 | (ax.toSet ++ adr)) ∧ (fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0] ⊦ ∀ adr. (adr ∈ fs0 | ax.toSet ↔ adr ∈ fs1 | ax.toSet) ∧ (fs0 | ax.toSet)[adr] = (fs1 | ax.toSet)[adr], fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax)

Rule argument

spec: store, inst: node-store name: Extension seq: ⊦ fs = fs0 ↔ (∀ adr. (adr ∈ fs ↔ adr ∈ fs0) ∧ fs[adr] = fs0[adr]) subst: [fs0, fs] → [fs1 | ax.toSet, fs0 | ax.toSet] flags: <all paths>

Additional information

Used terms: fs1 | ax.toSet, fs0 | ax.toSet