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Proof step 47 from proof keys-same-fs (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 48:
⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr]
Proof step 49:
(fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0], adr0 = adr ∨ adr0 ∈ ax → (adr0 ∈ fs0 ↔ adr0 ∈ fs1) ⊦ (∅)[adr0] = (fs1 | ax.toSet)[adr0], adr0 ∈ fs0 | ax.toSet, adr0 ∈ fs0 ∧ adr0 ∈ ax, fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax)

Rule argument

spec: store, inst: node-store name: notmemlookup seq: ⊦ ¬ adr ∈ fs → fs[adr] = (∅)[adr] subst: [adr, fs] → [adr0, fs0 | ax.toSet] flags: <all paths>

Additional information

Used simplifier rules: ⊦ adr ∈ fs | as ↔ adr ∈ fs ∧ adr ∈ as; (ws)% ⊦ adr ∈ ax.toSet ↔ adr ∈ ax; (s) Used terms: adr0, fs0 | ax.toSet