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Proof step 31 from proof keys-same-fs (previous step)

adr0 ∈ fs0 | (ax.toSet ++ adr) ↔ adr0 ∈ fs1 | (ax.toSet ++ adr), (fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0] ⊦ fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax), (adr0 ∈ fs0 | ax.toSet ↔ adr0 ∈ fs1 | ax.toSet) ∧ (fs0 | ax.toSet)[adr0] = (fs1 | ax.toSet)[adr0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ adr ∈ fs | as ↔ adr ∈ fs ∧ adr ∈ as
Proof step 33:
adr0 ∈ fs0 ∧ adr0 ∈ ax.toSet ++ adr ↔ adr0 ∈ fs1 | (ax.toSet ++ adr), (fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0] ⊦ (adr0 ∈ fs0 | ax.toSet ↔ adr0 ∈ fs1 | ax.toSet) ∧ (fs0 | ax.toSet)[adr0] = (fs1 | ax.toSet)[adr0], fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax)

Rule argument

spec: store+set, inst: node-store name: restrict-in seq: ⊦ adr ∈ fs | as ↔ adr ∈ fs ∧ adr ∈ as subst: [as, fs, adr] → [ax.toSet ++ adr, fs0, adr0] flags: <all paths>

Additional information

Used terms: ax.toSet ++ adr, fs0, adr0