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Proof step 39 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) ⊦ (adr0 ∈ fs0 ∧ adr0 ∈ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
(fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0], adr0 = adr ∨ adr0 ∈ ax → (adr0 ∈ fs0 ↔ adr0 ∈ fs1) ⊦ fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax), (adr0 ∈ fs0 ∧ adr0 ∈ ax ↔ adr0 ∈ fs1 | ax.toSet) ∧ (fs0 | ax.toSet)[adr0] = (fs1 | ax.toSet)[adr0]

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ ax.toSet ↔ adr ∈ ax; (s)