Semirings related

Question

How to prove a distributive near-semiring $(S,+,.)$ with multiplicative identity $1$ is additive subcommutative?

[ Hints: A non empty set $S$ with two binary operations $'+'$ and $'.'$ is called near-semirirng if the following conditions are satisfied:

$(i) a+(b+c)=(a+b)+c$

$(ii) a.(b.c)=(a.b).c$

$(iii) a.(b+c)=a.b+a.c$ $\forall$ a, b, c $\in S$ (left distributive law)

Distributive near-semiring means in addition with all above three conditions, $(a+b)c=ac+bc$ $\forall$ $a,b,c \in S $ (right distributive law) also satisfied.

A near-semiring with multiplicative identity means that there is an element $1 \in S$ such that $a.1=1.a=a$ $\forall $ $ a \in S$

A near-semiring $(S,+,.)$ is said to be additive subcommutative if $a+b+c+d=a+c+b+d$ $\forall a,b,c,d \in S$]

Answer 1

I tried to prove the question as in the following way:

$(1+1)(a+b)=1(a+b)+1(a+b)=a+b+a+b$ ....(1)(use right distributive law)\

Also $(1+1)(a+b)=(1+1)a+(1+1)b=a+a+b+b$...(2) (use left and right distributive laws)

From equations $(1)$ and $(2)$, we have $a+b+a+b=a+a+b+b$. Is it enough to prove additive subcommutative? Is it correct or not?