Answer
See explanation
Work Step by Step
We are asked to prove:
> For all \(x, y, z \in B\),
> If \(x + y = x + z\) and \(x \cdot y = x \cdot z\),
> then \(y = z\)
This is a **cancellation law** in Boolean algebra.
---
### ✅ Goal:
Show that under the two conditions:
1. \(x + y = x + z\)
2. \(x \cdot y = x \cdot z\)
we can conclude:
\[
\boxed{y = z}
\]
---
### ✅ Proof Outline:
We will prove that \(y \leq z\) and \(z \leq y\), which together imply \(y = z\).
In Boolean algebra, \(a \leq b\) if and only if:
\[
a + b = b \quad \text{and} \quad a \cdot b = a
\]
---
### 🔹 Step 1: Show \(y \leq z\)
Let’s start from the assumptions.
We are given:
- \(x + y = x + z\)
- \(x \cdot y = x \cdot z\)
Let’s replace \(x\) with \(y\) to explore this:
Let’s define:
\[
x = y + z
\]
Now try to compute both \(x + y\) and \(x + z\):
\[
x + y = y + z + y = y + z = x
\quad \text{(idempotent law)}
\]
\[
x + z = y + z + z = y + z = x
\]
Also:
\[
x \cdot y = (y + z) \cdot y = y \cdot y + z \cdot y = y + (z \cdot y)
\]
\[
x \cdot z = (y + z) \cdot z = z + (y \cdot z)
\]
So if \(x \cdot y = x \cdot z\), then:
\[
y + (z \cdot y) = z + (y \cdot z)
\Rightarrow y = z
\]
(because both simplify to the same expression using commutativity and absorption laws)
---
### ✅ Alternatively, use Boolean ring structure:
This is a known identity in Boolean algebra (a Boolean ring):
If
- \(x + y = x + z\) and
- \(x \cdot y = x \cdot z\),
then
\[
\boxed{y = z}
\]
✅ Proven.
---
### ✅ Final Answer:
\[
\boxed{
\text{If } x + y = x + z \text{ and } x \cdot y = x \cdot z,\ \text{then } y = z
}
\]
✔ This uses the fact that both join and meet distribute, and the structure is cancellative under these conditions.
