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If xzuy £ N, then xzuy G Dm for some m G /. Clearly m = i because xzHxzuy and m = k because uyCxzuy. This contradicts the choice of i ^ k and shows that xzuy G /V. Then (zxz)(uyu) G /V, and zxzlLz,uyulLu. Since zuHzxzuyu, from the  y is written x' (z e) y with a proper generic edge z. e. Lemma 4.2. Let z < e be a proper generic edge with e = (u, v) e E and let x, y e Irr(R). Then we have rop o g(xz e y) o q or ' if xz e Irr(R), 3 g(z e). y ~0 | p otherwise, (4.1) where r = pe(xzuy), D. 3.0 La différentielle du produit xzuy est zuydx—h xuy d z— hxzydu— hxzudy, en supposant que toutes les variables croissent en méme tems. On le démontre en faisant xx» — î, ou xzuy—sy, par conséquent zudx

ìxudzixzdu—dS).8c d._31 '>17 564mm?: euro 7171 115d' UZLUKUZl do H741; 6111 3m*: 6111178 11 am'momz: 7111* m1 T15(YU'ZUA, 1108*: 4x MVZ.>71 11'? do 31171171' 1171 119 11a ode511) :7271177 Ödor) a m1.: *xZuy xx 71111570710 v0 82 575 Multiply the Fluxion of each, by the Product of the rest of the Quantities, and the Sum of the Products thus arising will be the Fluxion sought *. , Thus the Fluxion of xy, is xy \yx ; that of xyz, is xyz+xzy fozx j and that of tcyzu, is xyzu\xyuz\xzuy Proposition 5.4 In any open coherent category the recognizable subobject classifier 1 + 1 together with bo, b1, not, and, and or forms a boolean algebra such that: • (x2, xv).and = xeny, O (xx, xy).or = Xzuy, •

Xr.not.= Xr, • 1x.bo = xix, • 1x.b1  y) I 4x on the curve y I émafl (b) u(z, y) I m3 on the curve 33/2 I 211:3 (c) u(.r,y) I sin(a:) on the line y I 0 6. yzu;c + xzuy I g2 (a) u(x, y) I an on y I 4a: (b) 14ml) = —2y On a3 = $3 — 2 (C) “($41) = 112 On y = —w 1.4 The QuasiLinear Equation Thusthe fluxion of xy, is xy +7* that of jfti is xyz + xzy + yzx ; and that of xyzu, is xyzu j xyuz + xzuy f yzux. (22.) Rule III. To find the fluxion of a fraction. From the fluxion of the numerator multiplied by the denominator, subtract the fluxion of the 0, (b) x2 Uj + y2 uy + z(x + y) u2 = 0, (c) x (y z) ux + y {z x) uy + z (x y) u, = 0, (d) yz ux xzuy + xy (x2 + y2) uz = 0, (e) x (y2 z2) ux+y (z2 y2) u„ + « (x2 y2)

«2.= 0. 9. Solve the equation ux + x uy = y with the Cauchy data (a) u(0,y) = y2.2.6 A 3partite tournament x z u y X Y Z a b x z u y X Y Z Conversely, suppose that T contains a directed cycle C that in turn contains vertices from at least three partite sets. Assume that C has the least possible length. Then there exist three