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Logical Effort calculation with multiple inputs

ben1122

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I saw this exercise during studying for my test in digital electronics.

1740946267275.png


At answers, they said that the logical effort at stage 4 (NOR3) is basically 7/3 and the branch (b) at stage 3 is b=3/2.
The branch at stage 3 is obvious - b_3 = 3/2

But the problem is with the logical effort, why isn't the logical effort at stage 4 is:
7/3 and not 14/3?

Note that logical effort is:
LE=g=C_g_GATE / C_g_ref
As C_g_ref of inverter is 3, i.e.:
C_g_ref=3
 
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