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Proof step 29 from proof index-update-adr-in (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: key abort: F discard: T computed: T

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];% ⊦ key ∈ ri[key, adr]; (s)% ⊦ ri[key, adr][key] = adr; (s) Used terms: key