Textbook
For propositions $P$ and $Q$, the conjunction $P\land Q$ is true exactly when both propositions are true. Every other combination gives false. The symbol $\land$ is read as “and.”
When the conditions depend on $x$, both concern the same chosen $x$. For an integer, “even and greater than 3” holds for 4 but not for 2 or 5. The formula specifies a truth value. Evaluation order and short-circuiting in a program belong to that language’s contract.
Conjunction depends only on its parts’ truth values: exchanging $P$ and $Q$ leaves the result unchanged. Proving $P\land Q$ requires justifying each part. Refuting it requires only one false part. When a combined condition concerns every element of a set, separate successful examples for its separate parts do not replace checking one arbitrary element against both.
Definitions and further reading: Mathematics for Computer Science.
In plain language
Conjunction requires both conditions, like a parcel rule demanding identification and a matching code. One person’s identification and another person’s code do not satisfy both requirements for a single recipient. The order of checks leaves the logical answer unchanged, while one failure makes it false. An attractive passport cannot replace a missing code; the everyday model fixes the subject of the checks without specifying a program’s execution order.
Example
The illustrative values 2, 5, and 4 are tested against “even and greater than 3.” Each step shows both checks and the combined answer. In the formulas, 1 denotes true and 0 denotes false.
Conjunction: Both Conditions Must Hold
Value 2
Even but not greater than three: false.
Step 1 of 3: Value 2
Read all steps
Value 2
Even but not greater than three: false.
- State
- P: truecurrentQ: falsecurrentfalsecurrent
Value 5
Greater than three but odd: false.
- State
- P: falsecurrentQ: truecurrentfalsecurrent
Value 4
Both conditions hold: true.
- State
- P: truecurrentQ: truecurrenttruecurrent
The numbers 2 and 5 illustrate different failures: one misses the threshold, the other misses evenness. Their successful halves cannot be pooled into a claim that the collection “passed on average.” For 4, both answers concern the same number, so the combined result is true. The unshown combination with two false parts also gives false; for instance, 1 satisfies neither condition.
Related concepts: Predicates.