We first observe that NP-completeness immediately follows forCNFSAT and SAT since 3SAT ≤p CNFSAT ≤p SAT . Note that 3SAT ⊂ CNFSAT ⊂ SAT . The reduction in each case is almost the identity (though of course it can’t be since the sets are different). The only issue is that we have to modify satisfiable formulas that are not in 3CNF. For the 3SAT ≤p CNFSAT reduction we simply map any input that is...