Construct a truth table for the statement (p<--->q)--->p
For the double arrow:
Φ Ψ Φ ↔ Ψ
T T T
T F F
F T F
F F T
For the arrow:
Φ Ψ Φ → Ψ
T T T
T F F
F T T
F F T
so putting them together,
p q result
T T T
T F F
F T T
F F T
check all that.
To construct a truth table for the statement "(p<--->q)--->p," we need to consider all possible combinations of the truth values of p and q. The truth values of p and q can either be true (T) or false (F).
Let's start by listing all possible combinations of p and q in the leftmost columns of the truth table:
p | q |
------
T | T |
T | F |
F | T |
F | F |
Now, let's determine the truth value of "(p<--->q)" for each combination. The "<--->" symbol represents the biconditional operator, which is true if and only if p and q have the same truth value.
For the first row, p=T and q=T. Since p and q have the same truth value, "(p<--->q)" is true (T). We fill in the first row of the truth table accordingly.
p | q | (p<--->q) |
-----------------
T | T | T |
T | F | F |
F | T | F |
F | F | T |
Now, let's evaluate "(p<--->q)--->p" for each combination of p and q. The "--->" symbol represents the implication operator, which is true unless the first statement is true and the second statement is false.
For the first row, "(p<--->q)" is true (T), and p is also true (T). Therefore, "(p<--->q)--->p" is true (T). We fill in the truth value in the last column for the first row.
p | q | (p<--->q) | (p<--->q)--->p |
---------------------------------
T | T | T | T |
T | F | F | T |
F | T | F | T |
F | F | T | F |
As a result, we have obtained the full truth table for the statement "(p<--->q)--->p."