First, note that union corresponds to max(a,b), and intersection to min(a,b).

Then note these identities:

abs(x) = pos(x) + pos(-x) abs(a - b) = pos(a - b) + pos(-a + b) a + b + abs(a - b) = 2*max(a,b) a + b - abs(a - b) = 2*min(a,b)So, for example:

[ r1 ] [ 1 1 1 ] [ pos[x1] ] [ 1 1 ] [ a ] [ r2 ] = [ 1 -1 -1 ] [ pos[x2] ] [ 1 -1 ] [ b ] [ pos[x3] ] [ -1 1 ]expands to:

r1 = a + b + pos(a - b) + pos(-a + b) = 2*max(a,b) r2 = a + b - pos(a - b) - pos(-a + b) = 2*min(a,b)These are our two functions we want to reproduce:

set-union(f,g): [max(f[k],g[k]) for k in range(len(f))] set-intersection(f,g): [min(f[k],g[k]) for k in range(len(f))]And here they are in MatSumSig (ignoring a factor of 2):

[ U1 ] [ 1 1 1 0 0 0 0 0 0 0 0 0 ] [ pos[x1] ] [ 1 1 0 0 0 0 0 0 ] [ f1 ] [ I1 ] = [ 1 -1 -1 0 0 0 0 0 0 0 0 0 ] [ pos[x2] ] [ 1 -1 0 0 0 0 0 0 ] [ g1 ] [ U2 ] [ 0 0 0 1 1 1 0 0 0 0 0 0 ] [ pos[x3] ] [ -1 1 0 0 0 0 0 0 ] [ f2 ] [ I2 ] [ 0 0 0 1 -1 -1 0 0 0 0 0 0 ] [ pos[x4] ] [ 0 0 1 1 0 0 0 0 ] [ g2 ] [ U3 ] [ 0 0 0 0 0 0 1 1 1 0 0 0 ] [ pos[x5] ] [ 0 0 1 -1 0 0 0 0 ] [ f3 ] [ I3 ] [ 0 0 0 0 0 0 1 -1 -1 0 0 0 ] [ pos[x6] ] [ 0 0 -1 1 0 0 0 0 ] [ g3 ] [ U4 ] [ 0 0 0 0 0 0 0 0 0 1 1 1 ] [ pos[x7] ] [ 0 0 0 0 1 1 0 0 ] [ f4 ] [ I4 ] [ 0 0 0 0 0 0 0 0 0 1 -1 -1 ] [ pos[x8] ] [ 0 0 0 0 1 -1 0 0 ] [ g4 ] [ pos[x9] ] [ 0 0 0 0 -1 1 0 0 ] [ pos[x10] ] [ 0 0 0 0 0 0 1 1 ] [ pos[x11] ] [ 0 0 0 0 0 0 1 -1 ] [ pos[x12] ] [ 0 0 0 0 0 0 -1 1 ]which expands to:

[U1,U2,U3,U4] = 2* [max(f1,g1), max(f2,g2), max(f3,g3), max(f4,g4)] [I1,I2,I3,I4] = 2* [min(f1,g1), min(f2,g2), min(f3,g3), min(f4,g4)]So I guess you could say they are space based union and intersection. We can also do a time based one:

[ U[t] ] = [ 1 1 1 ] [ pos[x1] ] [ 1 1 ] [ f ] [ I[t] ] [ 1 -1 -1 ] [ pos[x2] ] [ 1 -1 ] [ g ] [ pos[x3] ] [ -1 1 ]where U[t] is union of f[t] and g[t] with respect to time, and I[t] is intersection.

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