First, let's understand the operations involved in the question.

1. RoS (R composition S) is a relation from S to R such that for any elements x, y, z, if (x, y) is in S and (y, z) is in R, then (x, z) is in RoS.

2. (SoR)^-1 is the inverse of the relation SoR. If (x, y) is in SoR, then (y, x) is in (SoR)^-1.

Now, let's calculate RoS and (SoR)^-1.

RoS = {(x, z) | there exists a y such that (x, y) is in S and (y, z) is in R}
= {(a, b), (a, d), (b, c), (d, a)}

|RoS| = 4

SoR = {(x, z) | there exists a y such that (x, y) is in R and (y, z) is in S}
= {(b, d), (d, c), (a, a)}

(SoR)^-1 = {(d, b), (c, d), (a, a)}

|(SoR)^-1| = 3

Therefore, |RoS| + |(SoR)^-1| = 4 + 3 = 7.

Question

First, let's understand the operations involved in the question.

1. RoS (R composition S) is a relation from S to R such that for any elements x, y, z, if (x, y) is in S and (y, z) is in R, then (x, z) is in RoS.

2. (SoR)^-1 is the inverse of the relation SoR. If (x, y) is in SoR, then (y, x) is in (SoR)^-1.

Now, let's calculate RoS and (SoR)^-1.

RoS = {(x, z) | there exists a y such that (x, y) is in S and (y, z) is in R}
     = {(a, b), (a, d), (b, c), (d, a)}

|RoS| = 4

SoR = {(x, z) | there exists a y such that (x, y) is in R and (y, z) is in S}
     = {(b, d), (d, c), (a, a)}

(SoR)^-1 = {(d, b), (c, d), (a, a)}

|(SoR)^-1| = 3

Therefore, |RoS| + |(SoR)^-1| = 4 + 3 = 7.

Knowee AI · Accepted Answer