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Proof step 34 from proof index-update-adr-in (previous step)

# kx0 = # ax1, key0 ∈ ri ⊦ key0 = key, dups(kx0), ri[key0] ∈ ri[key, adr1][kx0, ax1], key ∈ kx0, ri[key0] ∈ ax1, ri[key0] ∈ ri[key, adr1] \ kx0.toSet, key0 ∈ kx0, key0 ∈ kx0.toSet

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
# kx0 = # ax1, key0 ∈ ri ⊦ key0 = key, dups(kx0), ri[key0] ∈ ri[key, adr1] \ kx0.toSet, ri[key0] ∈ ax1, key ∈ kx0, ri[key0] ∈ ri[key, adr1][kx0, ax1], key0 ∈ kx0

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ kx.toSet ↔ key ∈ kx; (s)