Answer
See explanation
Work Step by Step
We are now tracing **Algorithm 6.1.1** with:
- \(m = 4\), \(n = 4\)
- Set \(A = \{u, v, w, x\}\)
→ \(a[1]=u,\; a[2]=v,\; a[3]=w,\; a[4]=x\)
- Set \(B = \{r, u, y, z\}\)
→ \(b[1]=r,\; b[2]=u,\; b[3]=y,\; b[4]=z\)
---
## Initial values:
```text
i := 1
answer := "A ⊆ B"
```
---
## 🔁 First iteration (i = 1)
- `a[1] = u`
**Inner loop: look for u in b[1..4]:**
- j = 1 → b[1] = r ≠ u → found = "no"
- j = 2 → b[2] = u = u → found := "yes"
✅ `found = yes` → continue
Set `i := 2`
---
## 🔁 Second iteration (i = 2)
- `a[2] = v`
**Inner loop: look for v in b[1..4]:**
- j = 1 → b[1] = r ≠ v → found = "no"
- j = 2 → b[2] = u ≠ v
- j = 3 → b[3] = y ≠ v
- j = 4 → b[4] = z ≠ v
All checked → still `found = "no"`
So update:
```text
answer := "A ⊈ B"
```
Since answer has changed, the outer loop **terminates** early.
## ✅ Final Output:
\[
\boxed{\text{A ⊈ B}}
\]
(because \(v \notin B\))
