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Proof step 17 from proof index-update-in (
previous step
)
# kx0 = # ax0, ∀ ax, key, ri. # ax0 = # ax → (key ∈ ri[kx0, ax] ↔ key ∈ ri ∨ key ∈ kx0) ⊦ key0 ∈ ri, key0 ∈ ri[key0, adr][kx0, ax0]
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: ax0, key0, ri[key0, adr]
Additional information
Used simplifier rules: ⊦ key ∈ ri[key, adr]; (s) Used terms: ax0, key0, ri[key0, adr]