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:

- Reflexivity: For all x ∈ A, (x, x) ∈ R1 R2
- Symmetry: For all x, y ∈ A, if (x, y) ∈ R1 R2, then (y, x) ∈ R1 R2
- Transitivity: For all x, y, z ∈ A, if (x, y) ∈ R1 R2 and (y, z) ∈ R1 R2, then (x, z) ∈ R1 R2

Proof:

- 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.
- 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.
- 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.