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Proof step 12 from proof index-update-at-last-oc (
previous step
)
# ax0 = # kx + # kx1 + 1, ∀ ax, kx1, ri, key. # ax = # kx + # kx1 + 1 → ri[kx + key ' + kx1, ax][key] = ax[# kx] ∨ key ∈ kx1 ⊦ ri[key0, adr][kx + key ' + kx1, ax0][key] = ax0[# kx], key ∈ kx1
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: ax0, kx1, ri[key0, adr], key
Additional information
Used terms: ax0, kx1, ri[key0, adr], key