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Proof step 28 from proof keys-same-fs (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
∀ adr0. (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]

Rule argument

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