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Proof step 4 from proof index-update-adr-in (previous step)

#(key + kx0) = # ax, ∀ ax, adr, ri. # kx0 = # ax → (adr ∈ ri[kx0, ax] ↔ adr ∈ ax ∨ adr ∈ ri \ kx0.toSet) ∨ dups(kx0) ⊦ adr ∈ ri[key + kx0, ax] ↔ adr ∈ ax ∨ adr ∈ ri \ (key + kx0).toSet, dups(key + kx0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# kx0 + 1 = # ax, ∀ ax0, adr0, ri0. # kx0 = # ax0 → (adr0 ∈ ri0[kx0, ax0] ↔ adr0 ∈ ax0 ∨ adr0 ∈ ri0 \ kx0.toSet) ⊦ adr ∈ ri[key ' + kx0, ax] ↔ adr ∈ ax ∨ adr ∈ ri \ (kx0.toSet ++ key), dups(kx0), key ∈ kx0

Rule argument

Additional information

Used simplifier rules: ⊦ dups(key ' + kx) ↔ dups(kx) ∨ key ∈ kx; (ws)%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)