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Proof step 4 from proof range-in-ri-keys-in-ri (previous step)

ri-cons(ri, fs), (adr + ax0).toSet ⊆ range ri, ∀ ri, fs. ri-cons(ri, fs) ∧ ax0.toSet ⊆ range ri → keys(fs, ax0).toSet ⊆ dom ri ⊦ keys(fs, adr + ax0).toSet ⊆ dom ri

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
ri-cons(ri, fs), adr ∈ ri, ax0.toSet ⊆ range ri, ∀ ri, fs. ri-cons(ri, fs) ∧ ax0.toSet ⊆ range ri → keys(fs, ax0).toSet ⊆ dom ri ⊦ keys(fs, ax0).toSet ++ fs[adr].key ⊆ dom ri

Rule argument

Additional information

Used simplifier rules: 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)% ⊦ keys(fs, adr ' + ax) = fs[adr].key + keys(fs, ax); (s)% ⊦ adr ∈ range ri ↔ adr ∈ ri; (s)% ⊦ as ++ adr ⊆ as0 ↔ as ⊆ as0 ∧ adr ∈ as0; (ws)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]%unc: ⊦ as ∪ as0 = as0 ∪ as; from specification set-union[node-store]%una: ⊦ (as ∪ as0) ∪ as1 = as ∪ as0 ∪ as1; from specification set-union[node-store]% ⊦ as ∪ ∅ = as; (s)% ⊦ (as ++ adr) ∪ as0 = (as ∪ as0) ++ adr; (s)% ⊦ adr '.toSet = ∅ ++ adr; (s)% ⊦ (ax + ax0).toSet = ax.toSet ∪ ax0.toSet; (s)% ⊦ adr + ax = adr ' + ax; (s)