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Proof step 23 from proof index-update-adr-in (previous step)

# kx0 = # ax1, key0 ∈ ri, ri[key0] ∈ ri[key, adr1][kx0, ax1] ⊦ ri[key0] = adr1, key0 = key, dups(kx0), key0 ∈ kx0, key ∈ kx0, ri[key0] ∈ ax1, ∃ key1. key1 ≠ key ∧ key1 ∈ ri \ kx0.toSet ∧ (ri \ kx0.toSet)[key1] = ri[key0]

User applied rule exists right

And closed the goal!

Rule argument

pos: right 7 terms: key0 abort: F discard: T computed: F

Additional information

Used simplifier rules: toSet-diff-in: ⊦ ¬ key ∈ kx → (key ∈ ri \ kx.toSet ↔ key ∈ ri);%toSet-diff-get: ⊦ ¬ key ∈ kx → (ri \ kx.toSet)[key] = ri[key]; Used terms: key0