7 Finally, we need to consider the scalar bidoublet and the scalar left doublet. In this model, ( φ 0 $_{1}$, φ − 1 ) will be heavy, and the two doublets ( φ + $_{2}$, φ 0 $_{2}$), ( φ + $_{L}$, φ 0 $_{L}$) are similar to the usual two Higgs doublets considered in the SM. The linear combination ( v 3 Φ 2 − v 2 Φ$_{L}$) / √ v 2 2 + v 2 3 = [ H + $_{L}$, ( H 0 $_{L}$+ iA 0 $_{L}$) $_{/}$√$_{2] is physical and light.}$ Since e has L = 1 and n has L = 0, the W + R of this model must have L = S − T 3 R = 0 − 1 = − 1. This also means that unlike the conventional LRM, W ± R does not mix with the W ± L of the SM at all. This important property allows the SU (2)$_{R}$ breaking scale to be much lower than it would be otherwise, as explained already 22 years ago [2, 3]. Assuming that g$_{L}$ = g$_{R}$ and let x ≡ sin $^{2}$θ$_{W}$ , then the neutral gauge bosons of the DLRM (as well as the ALRM) are given by with the coupling e/ √ x (1 − x )(1 − 2 x ). The masses of the gauge bosons are given by Whereas Z couples to the current J 3 L − xJ$_{em}$ with coupling e/ √ x (1 − x ) as in the SM, Z ′ couples to the current where zero Z − Z ′ mixing has been assumed, using the condition [3] v 2 $_{2}$/ ( v 2 $_{2}$+ v 2 $_{3}$) = x/ (1 − x ). Note that in the ALRM, Δ$_{R}$ is absent, hence v 6 = 0 in the above. Also, the assignment of ( ν, e )$_{L}$ there is different, hence the Z ′ of the DLRM is not identical to that of the ALRM. At the LHC, if a new Z ′ exists which couples to both quarks and leptons, it will be discovered with relative ease. Once M$_{Z}$ ′ is determined, then the DLRM predicts the existence of W ± R with a mass in the range (1 − 2 x ) 2(1 x ) M 2 Z ′ + x 2(1 − x ) $_{2}$M 2 W$_{L}$ < M 2 W$_{R}$ < (1 − 2 x ) (1 x ) M 2 Z ′ + x 2 (1 − x ) $_{2}$M 2 $_{W$_{L}$}$. (26) M 2 Z ′ = e $^{2}$(1 − x ) 2 x (1 2 x ) ( v 2 2 + v 2 4 + 4 v 2 $_{6}$) − x $^{2}$M2 W$_{L}$ (1 − x )(1 − 2 x ) , (25) M 2 W$_{L}$ = e 2 2 x ( v 2 2 + v 2 $_{3}$) , M 2 Z = M 2 W$_{L}$ 1 − x , M 2 W$_{R}$ = e 2 2 x ( v 2 2 + v 2 4 + 2 v 2 $_{6}$) , (24) J$_{Z}$ ′ = xJ 3 L + (1 − x ) J 3 R − xJ$_{em}$ (23)   $^{A}$ Z Z ′    =    √$_{x}$ √$_{x}$ √$_{1}$ − 2 x √$_{1}$ − x − $_{x/}$√$_{1}$ − x − √ x (1 − 2 x ) / (1 − x ) 0 √ (1 − 2 x ) / (1 − x ) − √ x/ (1 − x )      $^{W}$$^{0}$ L W 0 R B   $^{.}$ (22) IV. GAUGE SECTOR