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Proof step 2 from proof index-update-dom (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# kx0 + 1 = # ax, ∀ ax, ri. # kx0 = # ax → dom ri[kx0, ax] = dom ri ∪ kx0.toSet ⊦ dom ri[key ' + kx0, ax] = (kx0.toSet ∪ dom ri) ++ key

Rule argument

Additional information

Used simplifier rules: unc: ⊦ ks ∪ ks0 = ks0 ∪ ks; from specification set-union[node-index]%una: ⊦ (ks ∪ ks0) ∪ ks1 = ks ∪ ks0 ∪ ks1; from specification set-union[node-index]% ⊦ ks ∪ ∅ = ks; (s)% ⊦ (ks ++ key) ∪ ks0 = (ks ∪ ks0) ++ key; (s)% ⊦ key '.toSet = ∅ ++ key; (s)% ⊦ (kx + kx0).toSet = kx.toSet ∪ kx0.toSet; (s)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]% ⊦ m +1 = m + 1; (s)% ⊦ #(key ' + kx) = (# kx)+1; (s)% ⊦ key + kx = key ' + kx; (s)