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Proof step 17 from proof keys-same-fs (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
adr ∈ fs0 ↔ adr ∈ fs1 | (ax.toSet ++ adr), (fs0 | (ax.toSet ++ adr))[adr] = (fs1 | (ax.toSet ++ adr))[adr], keys(fs0, ax) = keys(fs1, ax), fs0 | ax.toSet = fs1 | ax.toSet ⊦ fs0[adr].key = fs1[adr].key

Rule argument

Additional information

Used simplifier rules: ⊦ adr ∈ as ++ adr0 ↔ adr = adr0 ∨ adr ∈ as; (s)