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CSE 215
HOMEWORK 1
- Truth table for (A → B)→ [((B → C) ∧ ~C) → ~A]
- Statement A - Ring is in Jar A Statement B - Ring is not in Jar A Statement C - Ring is not in Jar C
We know only one of the statements is correct, and the other two are incorrect.
Case 1: If Statement A is true, Statement B would be false - the ring is in Jar A. Statement C would be false - the ring is in Jar C. Therefore, Statement A and C are contradicting each other as each says the ring is in the respective jars. And hence, Statement A cannot be true.
Case 2: If Statement B is true, Statement A would be false - the ring is not in Jar A. Statement C would be false - the ring is in Jar C. Therefore, the ring is in Jar C, as Statement A and B support each other.
Case 3: If Statement C is true, Statement A would be false - the ring is not in Jar A. Statement B would be false - the ring is in Jar A. Therefore, Statement A and B are contradicting each other. And hence, Statement C cannot be true.
A B C ~A ~C A→B B→C (B→C) ∧ ~C (B→C) ∧ ~C)→~A (A→B)→[((B→C) ∧ ~C)→~A]
T T T F F T T F T T
T T F F T T F F T T
T F T F F F T F T T
T F F F T F T T F T
F T T T F T T F T T
F T F T T T F F T T
F F T T F T T F T T
F F F T T T T T T T
CSE 215
- Truth table for p ∧ q ∨ (~p ∧ ~q) and p ↔ q
p q ~p ~q p ∧ q ~p ∧ ~q p ∧ q ∨ (~p ∧ ~q) p ↔ q
T T F F T F T T
T F F T F F F F
F T T F F F F F
F F T T F T T T
Since “ p ∧ q ∨ (~p ∧ ~q) ” and “ p ↔ q ” have the same truth values they are logically equivalent.
- Assuming, ((p1→q) ∧ (p2→q)) is True, we must prove ((p1 ∨ p2) → q) is true as well. If ((p1→q) ∧ (p2→q)) is True, then (p1→q) and (p2→q) are also both true. Since, (p1→q) is True, then if p1 is true then q must also be True and similarly, (p2→q) is True, then if p2 is true then q must also be True. Furthermore, since (p1 ∨ p2) is True, then p1 and p2 must also be True, thus the above assumption is correct. Concluding, since (p1 ∨ p2) is True, then q must also be true. This implies that ((p1 ∨ p2) → q) is true. Hence, ((p1→q) ∧ (p2→q)) and ((p1 ∨ p2) → q) are logically equivalent since they have the same truth values.
- U: None of us is a knight. V: At least three of us are knights. W: At most three of us are knights. X: Exactly five of us are knights. Y: Exactly two of us are knights. Z: Exactly one of us is a knight.
Case 1: If U is telling the truth, then there are no knights that means even U cannot be a knight. Hence, U is lying and is a knave.
Case 2: If V is telling the truth then U, W, Y, and Z are lying and are knaves. That means less than 3 people are knights which contradicts V’s statement.
Case 3: If W is telling the truth then U, W, and X are lying and are knaves. That means either Y or Z are telling the truth thus less than 3 are knights - supporting W’s statement.
CSE 215
q ∨ ~r : I was cooking breakfast or I did not leave my glasses on the kitchen table. ~p : I was not reading the newspaper.
From premise 3 we can say that if r is false then in statement 1 p must also be false for a false r, hence the conclusion should be that p is false. Therefore, it is a valid statement.
- A. Truth table for A ∧ (A ∨ B) and A:
A B A ∨ B A ∧ (A ∨
B)
A
T T T T T
T F T T T
F T T F F
F F F F F
Since “ A ∧ (A ∨ B) ” and “ A ” have the same truth values they are logically equivalent.
B. Truth table for ~(A ∨ ~B) → ~B and B → A:
A B ~B A ∨ ~B ~(A ∨ ~B) ~(A ∨ ~B) →
~B
B → A
T T F T F T T
T F T T F T T
F T F F T F F
F F T T F T T
Since “ ~(A ∨ ~B) → ~B ” and “ B → A ” have the same truth values they are logically equivalent.
C. Truth table for (P ∧ (P → Q)) → Q and P → (P ↔ Q):
P Q P → Q P ∧ (P → Q) (P ∧ (P → Q)) → Q P ↔ Q P → (P ↔ Q)
T T T T T T T
T F F F T F F
F T T F T F T
CSE 215
F F T F T T T
Since “ (P ∧ (P → Q)) → Q ” and “ P → (P ↔ Q) ” do not have the same truth values; they are not logically equivalent.
D. Truth table for ~(~P ∧ Q) ∨ (P ∨ R) ∨ (Q ∧ ~R) and True:
Since “ ~(~P ∧ Q) ∨ (P ∨ R) ∨ (Q ∧ ~R) ” and “ True ” have the same truth values; they are not logically equivalent.
- The statement is True.
P Q R ~P ~R ~P ∧ Q ~(~P ∧ Q) P ∨ R Q ∧ ~R (~P ∧ Q) ∨ (P ∨ R) ∨ (Q ∧ ~R True
T T T F F F T T F T T
T T F F T F T T T T T
T F T F F F T T F T T
T F F F T F T T F T T
F T T T F T F T F T T
F T F T T T F F T T T
F F T T F F T T F T T
F F F T T F T F F T T
