Answer :
To determine which statements are tautologies, we need to analyze each given statement and check if it is always true, regardless of the truth values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
1. Statement 1: [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex]
- Let's break this statement down:
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- [tex]\( p \rightarrow q \)[/tex]: This is an implication, which is true in every case except when [tex]\( p \)[/tex] is true and [tex]\( q \)[/tex] is false.
- The disjunction ([tex]\(\vee\)[/tex]) operator means the entire expression [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex] is true if at least one of the parts is true.
- Upon examining all possible truth values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex], this statement is always true.
2. Statement 2: [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex]
- Let's break this statement down:
- [tex]\( p \wedge q \)[/tex]: This is true only if both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are true.
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- The implication ([tex]\(\rightarrow\)[/tex]) operator means the entire expression [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex] is true if [tex]\( p \wedge q \)[/tex] being true also implies that [tex]\( p \vee q \)[/tex] is true.
- If [tex]\( p \wedge q \)[/tex] is true, [tex]\( p \vee q \)[/tex] is necessarily true because both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are true. If [tex]\( p \wedge q \)[/tex] is false, the implication is true regardless of the truth value of [tex]\( p \vee q \)[/tex].
- Hence, this statement is always true.
3. Statement 3: [tex]\(q \rightarrow (p \wedge q)\)[/tex]
- Let's break this statement down:
- [tex]\( q \rightarrow (p \wedge q)\)[/tex]: This is an implication, which is true if [tex]\( q \)[/tex] implies [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
- If [tex]\( q \)[/tex] is true, then for the implication to be true, [tex]\( p \wedge q \)[/tex] must also be true, meaning [tex]\( p \)[/tex] must be true.
- If [tex]\( q \)[/tex] is false, the implication [tex]\( q \rightarrow (p \wedge q) \)[/tex] is always true regardless of [tex]\( p \)[/tex].
- Testing the case where [tex]\( q \)[/tex] is true and [tex]\( p \)[/tex] is false, we find that [tex]\( q \rightarrow (p \rightarrow q) \)[/tex] is false.
- Hence, this statement is not always true.
4. Statement 4: [tex]\((p \vee q) \rightarrow p\)[/tex]
- Let's break this statement down:
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- The implication ([tex]\(\rightarrow\)[/tex]) operator means the entire expression [tex]\((p \vee q) \rightarrow p\)[/tex] is true if [tex]\( p \vee q \)[/tex] being true also implies that [tex]\( p \)[/tex] is true.
- There are cases where [tex]\( q \)[/tex] is true and [tex]\( p \)[/tex] is false, making [tex]\( p \vee q \)[/tex] true but the overall implication false because [tex]\( p \)[/tex] is false.
- Hence, this statement is not always true.
From the analysis above, the two statements that are always true (tautologies) are:
1. [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex]
2. [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex]
These are the correct answers.
1. Statement 1: [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex]
- Let's break this statement down:
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- [tex]\( p \rightarrow q \)[/tex]: This is an implication, which is true in every case except when [tex]\( p \)[/tex] is true and [tex]\( q \)[/tex] is false.
- The disjunction ([tex]\(\vee\)[/tex]) operator means the entire expression [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex] is true if at least one of the parts is true.
- Upon examining all possible truth values of [tex]\( p \)[/tex] and [tex]\( q \)[/tex], this statement is always true.
2. Statement 2: [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex]
- Let's break this statement down:
- [tex]\( p \wedge q \)[/tex]: This is true only if both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are true.
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- The implication ([tex]\(\rightarrow\)[/tex]) operator means the entire expression [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex] is true if [tex]\( p \wedge q \)[/tex] being true also implies that [tex]\( p \vee q \)[/tex] is true.
- If [tex]\( p \wedge q \)[/tex] is true, [tex]\( p \vee q \)[/tex] is necessarily true because both [tex]\( p \)[/tex] and [tex]\( q \)[/tex] are true. If [tex]\( p \wedge q \)[/tex] is false, the implication is true regardless of the truth value of [tex]\( p \vee q \)[/tex].
- Hence, this statement is always true.
3. Statement 3: [tex]\(q \rightarrow (p \wedge q)\)[/tex]
- Let's break this statement down:
- [tex]\( q \rightarrow (p \wedge q)\)[/tex]: This is an implication, which is true if [tex]\( q \)[/tex] implies [tex]\( p \)[/tex] and [tex]\( q \)[/tex].
- If [tex]\( q \)[/tex] is true, then for the implication to be true, [tex]\( p \wedge q \)[/tex] must also be true, meaning [tex]\( p \)[/tex] must be true.
- If [tex]\( q \)[/tex] is false, the implication [tex]\( q \rightarrow (p \wedge q) \)[/tex] is always true regardless of [tex]\( p \)[/tex].
- Testing the case where [tex]\( q \)[/tex] is true and [tex]\( p \)[/tex] is false, we find that [tex]\( q \rightarrow (p \rightarrow q) \)[/tex] is false.
- Hence, this statement is not always true.
4. Statement 4: [tex]\((p \vee q) \rightarrow p\)[/tex]
- Let's break this statement down:
- [tex]\( p \vee q \)[/tex]: This is true if at least one of [tex]\( p \)[/tex] or [tex]\( q \)[/tex] is true.
- The implication ([tex]\(\rightarrow\)[/tex]) operator means the entire expression [tex]\((p \vee q) \rightarrow p\)[/tex] is true if [tex]\( p \vee q \)[/tex] being true also implies that [tex]\( p \)[/tex] is true.
- There are cases where [tex]\( q \)[/tex] is true and [tex]\( p \)[/tex] is false, making [tex]\( p \vee q \)[/tex] true but the overall implication false because [tex]\( p \)[/tex] is false.
- Hence, this statement is not always true.
From the analysis above, the two statements that are always true (tautologies) are:
1. [tex]\((p \vee q) \vee (p \rightarrow q)\)[/tex]
2. [tex]\((p \wedge q) \rightarrow (p \vee q)\)[/tex]
These are the correct answers.
