Show that if R1 and R2 are equivalence relations on A, then R1 R2 is an equivalence relation on A.

To show that R1 R2 is an equivalence relation on A, we need to prove that it satisfies the three properties of an equivalence relation:

1. Reflexivity: For all x ∈ A, (x, x) ∈ R1 R2
2. Symmetry: For all x, y ∈ A, if (x, y) ∈ R1 R2, then (y, x) ∈ R1 R2
3. Transitivity: For all x, y, z ∈ A, if (x, y) ∈ R1 R2 and (y, z) ∈ R1 R2, then (x, z) ∈ R1 R2

Proof:

1. Reflexivity: Since R1 and R2 are equivalence relations on A, we have (x, x) ∈ R1 and (x, x) ∈ R2 for all x ∈ A. Therefore, (x, x) ∈ R1 R2 for all x ∈ A, and R1 R2 is reflexive.
2. Symmetry: Suppose (x, y) ∈ R1 R2. Then there exist elements a, b ∈ A such that (x, a) ∈ R1, (a, y) ∈ R1, (x, b) ∈ R2, and (b, y) ∈ R2. Since R1 is an equivalence relation, we have (a, x) ∈ R1 and (y, a) ∈ R1. Similarly, R2 implies (b, x) ∈ R2 and (y, b) ∈ R2. Therefore, we have (y, a) ∈ R1 and (x, b) ∈ R2. This implies (y, x) ∈ R2 R1. By similar argument, we can prove (y, x) ∈ R1 R2. Hence, R1 R2 is symmetric.
3. Transitivity: Suppose (x, y) ∈ R1 R2 and (y, z) ∈ R1 R2. Then there exist elements a, b, c, d ∈ A such that (x, a) ∈ R1, (a, y) ∈ R1, (y, b) ∈ R2, (b, z) ∈ R2, (x, c) ∈ R1, (c, y) ∈ R1, and (y, d) ∈ R2, (d, z) ∈ R2. Since R1 is an equivalence relation, we have (a, x) ∈ R1 and (y, a) ∈ R1, and (c, x) ∈ R1 and (y, c) ∈ R1. Similarly, R2 implies (b, y) ∈ R2 and (z, b) ∈ R2, and (d, y) ∈ R2 and (z, d) ∈ R2. Therefore, we have (a, x) ∈ R1, (y, a) ∈ R1, (z, b) ∈ R2, and (y, d) ∈ R2. This implies (x, d) ∈ R1 R2. By similar argument, we can prove (x, z) ∈ R1 R2. Hence, R1 R2 is transitive.

Therefore, R1 R2 satisfies all three properties of an equivalence relation and is an equivalence relation on A.