Question

determine if each symbolic statement is true or false when... p is true, q is true, and r is false; r → p; (p ∨ r) → (p ∨ q); ¬p ∧ ¬r; p → r; ¬p → (p ∨ r); p → ¬(p ∨ r); (p ∧ r) → (p ∧ q); ¬(p ∧ r)

Explanation:

Step1: Recall logical - operator rules

We know that for implication \(a
ightarrow b\) is false when \(a\) is true and \(b\) is false, and true otherwise. For disjunction \(a\vee b\) is true if at least one of \(a\) and \(b\) is true, and for conjunction \(a\wedge b\) is true only when both \(a\) and \(b\) are true. Also, \(
eg a\) is the opposite of the truth - value of \(a\). Given \(p = T\), \(q = T\), \(r = F\).

Step2: Evaluate \(r

ightarrow p\)
Since \(r = F\), by the rule of implication, \(r
ightarrow p\) is \(T\) (because when the antecedent is false, the implication is always true).

Step3: Evaluate \((p\vee r)

ightarrow(p\vee q)\)
First, \(p\vee r=T\vee F = T\), and \(p\vee q=T\vee T = T\). Then, since the antecedent \(p\vee r\) is \(T\) and the consequent \(p\vee q\) is \(T\), \((p\vee r)
ightarrow(p\vee q)\) is \(T\).

Step4: Evaluate \(

eg p\wedge
eg r\)
\(
eg p = F\) and \(
eg r = T\), so \(
eg p\wedge
eg r=F\wedge T = F\).

Step5: Evaluate \(p

ightarrow r\)
Since \(p = T\) and \(r = F\), \(p
ightarrow r\) is \(F\).

Step6: Evaluate \(

eg p
ightarrow(p\vee r)\)
\(
eg p = F\), \(p\vee r=T\vee F = T\). Since the antecedent \(
eg p\) is \(F\), \(
eg p
ightarrow(p\vee r)\) is \(T\).

Step7: Evaluate \(p

ightarrow
eg(p\vee r)\)
\(p\vee r=T\vee F = T\), \(
eg(p\vee r)=F\). Since \(p = T\) and the consequent \(
eg(p\vee r)\) is \(F\), \(p
ightarrow
eg(p\vee r)\) is \(F\).

Step8: Evaluate \((p\wedge r)

ightarrow(p\wedge q)\)
\(p\wedge r=T\wedge F = F\), \(p\wedge q=T\wedge T = T\). Since the antecedent \(p\wedge r\) is \(F\), \((p\wedge r)
ightarrow(p\wedge q)\) is \(T\).

Step9: Evaluate \(

eg(p\wedge r)\)
\(p\wedge r=T\wedge F = F\), so \(
eg(p\wedge r)=T\).

Answer:

1. \(r

ightarrow p\): TRUE

1. \((p\vee r)

ightarrow(p\vee q)\): TRUE

1. \(

eg p\wedge
eg r\): FALSE

1. \(p

ightarrow r\): FALSE

1. \(

eg p
ightarrow(p\vee r)\): TRUE

1. \(p

ightarrow
eg(p\vee r)\): FALSE

1. \((p\wedge r)

ightarrow(p\wedge q)\): TRUE

1. \(

eg(p\wedge r)\): TRUE