Proof of Rearranging Sums, Differences and Two-Term Combinations in a Real Vector Space

Throughout, every sum and scalar multiple formed lies in $V$ (Membership and the Zero Law of a Real Vector Space in Flat Form (i), Membership and the Zero Law of a Real Vector Space in Flat Form (ii)), so the axioms of Vector Space over a Field apply to it; we write $v+(-v)=0_{V}$ for $v\in V$ by The Zero Vector, Negatives and Differences in a Vector Space (ii) and $x+0_{V}=x$ for $x\in V$ by The Zero Vector, Negatives and Differences in a Vector Space (i).

(i): using Vector Space over a Field (i) for the first, second, fourth and fifth equalities and Vector Space over a Field (ii) for the third,

$$(a+b)+(c+d)=a+(b+(c+d))=a+((b+c)+d)=a+((c+b)+d)=a+(c+(b+d))=(a+c)+(b+d).$$

(ii): by The Zero Vector, Negatives and Differences in a Vector Space (iii), Elementary Identities in a Vector Space (vi) and (i),

$$(x+b)-(y+b)=(x+b)+\bigl((-y)+(-b)\bigr)=\bigl(x+(-y)\bigr)+\bigl(b+(-b)\bigr)=(x-y)+0_{V}=x-y.$$

(iii): in the same way, by The Zero Vector, Negatives and Differences in a Vector Space (iii), Elementary Identities in a Vector Space (vi) and (i),

$$(x+y)-(a+b)=(x+y)+\bigl((-a)+(-b)\bigr)=\bigl(x+(-a)\bigr)+\bigl(y+(-b)\bigr)=(x-a)+(y-b).$$

(iv): by The Zero Vector, Negatives and Differences in a Vector Space (iii), Vector Space over a Field (ii) and Vector Space over a Field (i),

$$(a+w)-a=(w+a)+(-a)=w+\bigl(a+(-a)\bigr)=w+0_{V}=w.$$

(v): by Vector Space over a Field (vii) applied to both terms, and then (i) applied to the vectors $tp,tq,sp',sq'$,

$$t(p+q)+s(p'+q')=(tp+tq)+(sp'+sq')=(tp+sp')+(tq+sq').$$

(vi): by Vector Space over a Field (viii), the hypothesis $t+s=1$ and Vector Space over a Field (vi),

$$tx+sx=(t+s)x=1x=x.$$