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Proof step 4 from proof in-read-in-dom (previous step)

adr ∈ ri[key + kx0], (key + kx0).toSet ⊆ dom ri, ∀ adr, ri. adr ∈ ri[kx0] ∧ kx0.toSet ⊆ dom ri → adr ∈ ri ⊦ adr ∈ ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
key ∈ ri, adr ∈ ri[kx0], kx0.toSet ⊆ dom ri, ∀ adr, ri. adr ∈ ri[kx0] ∧ kx0.toSet ⊆ dom ri → adr ∈ ri ⊦ adr = ri[key], adr ∈ ri

Rule argument

Additional information

Used simplifier rules: index-lookup-rec: ⊦ ri[key ' + kx] = ri[key] + ri[kx];%key ∈ ri ⊦ ri[key] ∈ ri; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ adr ∈ adr0 ' + ax ↔ adr = adr0 ∨ adr ∈ ax; (s)% ⊦ adr + ax = adr ' + ax; (s)% ⊦ key ∈ dom ri ↔ key ∈ ri; (s)% ⊦ ks ++ key ⊆ ks0 ↔ ks ⊆ ks0 ∧ key ∈ ks0; (ws)%a: ⊦ (kx + kx0) + kx1 = kx + kx0 + kx1; from specification list[key-list-set]%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)% ⊦ key + kx = key ' + kx; (s)