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Proof step 5 from proof index-update-dom (previous step)

# kx0 = # ax0, ∀ ax, ri. # ax0 = # ax → dom ri[kx0, ax] = dom ri ∪ kx0.toSet ⊦ dom ri[key, adr][kx0, ax0] = (kx0.toSet ∪ dom ri) ++ key

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: ax0, ri[key, adr]

Additional information

Used simplifier rules: ⊦ dom ri[key, adr] = dom ri ++ key; (s)% ⊦ (ks ++ key) ∪ ks0 = (ks ∪ ks0) ++ key; (s)%una: ⊦ (ks ∪ ks0) ∪ ks1 = ks ∪ ks0 ∪ ks1; from specification set-union[node-index]%unc: ⊦ ks ∪ ks0 = ks0 ∪ ks; from specification set-union[node-index] Used terms: ax0, ri[key, adr]