276 Wilder’s Algebraic Solution. 
+S8/1S’m—1 \ +S/1S’m—1 
+pSm—2 | +p'S'm—2 
yemnni) tpSm ag 2 \ mma) t q'S'm named 3 eae Sd 
+19m —(n—2 | +tS'm —(n—-2) | 
+-uSm—(n—1) ) +u? Sm —(n— 1) ) 
+8”’1S’m—1 4817 3VS™Im-1 ) g 
+p"S"m—2 4 pe sC0S—-80) m2 3 
+q7S'm— 3 | te sj gh US mS - 
shocker | fpr 3gr- 0 m-n-2 | F 
+u2S"m-(n-1) j y-18"-3()m—(n-1) J} © 
for the general expression of (A’); the accented letters hav- 
ing the same relation to a8-t-ay+-ad-+-By etc., aSy +80 
+ayb+B75 etc., a87i+etc., that $1, Sm—1, p, Sm—2, 
etc., have toa+S8+7-+6 elc., a—!-+6"—!-+y"™"1+0"~) etc., 
aBtayto+By etc., a52418"-2-+y"-2 -+0"—? etc. 
- Proposition second. Let o(aypq etc.), and ¢'(xypq. etc.), 
be two functions of the independents xypq etc., and let 9 be 
a factor of o’, then I say that if any function 9’(xpq etc.), 
written for y in 9’, makes it identically nothing, it will also, 
when written for y, make 9 identically nothing. For if not, 
we shall have by putting y’=9"(xpq etc.), 
'(xy’ pq etc.) 
o(xy'pq ete.) — o(xy'pq ete. 
hing. And, reciprocally, if ¢” when written for y in ¢, makes 
it identically nothing, it will when written for y in @, make © 
it identically nothing. For if not, we shall have 
o(xy'pq etc.) 9% (ry'pq ete. . 
fae ae s an ) or 0, a factor of ¢, which is 
oy . EL eH 
Proposition third. Tf the function ®”(xpq etc.), written for 
y, makes both @ and 9 identically nothing, then I say that 
~ al ?is.a ae of - #’ is a factor of >, for we either have 
(xy'pq ete.) 0 o(xry'pq etc.) 0 : ee 
aey’pq te.) 0% Fi sap ete )~0% 204 if the division 
does not take place independently of y’, or what is the same, ¥> 
we shall have a relation between 2, p, q, etc., which is contra- 
ty to the hypothesis. We either have therefore o(rypq etc.), 
factor of ¢(xypq etc.), or *(rypg etc.), a factor of o(zyPd 
9. 
yet nothing divisible by some- 
ete.), according as 9’ > or< 
