Hey Invincible, good to see you again.
Here are the solutions to your questions:
1) Define the following terms and give illustrations where necessary:
- a. Proposition: A declarative sentence that is either true or false, but not both.
- Illustration: "The sky is blue" is a proposition. "Is the sky blue?" is not.
- b. Fallacy: A mistaken belief or an error in reasoning that renders an argument invalid.
- Illustration: The argument "All cats have four legs. My dog has four legs. Therefore, my dog is a cat" is a fallacy of the undistributed middle.
- c. Contradiction: A compound proposition that is always false, regardless of the truth values of its constituent propositions.
- Illustration: P∧∼P is a contradiction.
- d. Contingency: A compound proposition that is neither a tautology nor a contradiction; its truth value depends on the truth values of its constituent propositions.
- Illustration: P∧Q is a contingency, as its truth value can be true or false depending on P and Q.
- e. Argument: A set of statements, one of which is designated as the conclusion and the others as premises, intended to provide support for the conclusion.
- Illustration: "All men are mortal. Socrates is a man. Therefore, Socrates is mortal" is an argument.
2) Verify whether each of the following statements is a tautology or contradiction:
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i) [(A⇔B)∨(C⇔B)]⇒∼A
This statement is a contingency.
- Explanation: If A is True, B is True, and C is True, then (T⇔T)∨(T⇔T) is T∨T, which is True. The implication becomes T⇒∼T, which is T⇒F, which is False. Since it can be false, it is not a tautology. If A is False, then ∼A is True, making the entire implication True regardless of the antecedent. Since it can be both true and false, it is a contingency.
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ii) (A∧B)∧∼(A∨B)
This statement is a contradiction.
- Explanation:
(A∧B)∧∼(A∨B)
Using De Morgan's Law, ∼(A∨B)≡(∼A∧∼B):
(A∧B)∧(∼A∧∼B)
Rearranging terms (associativity and commutativity of ∧):
(A∧∼A)∧(B∧∼B)
Since A∧∼A is always False and B∧∼B is always False:
F∧F
F
The statement is always false.
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iii) [(A⇒B)∧A]⇒A
This statement is a tautology.
- Explanation:
[(A⇒B)∧A]⇒A
Using the equivalence P⇒Q≡∼P∨Q:
∼[(∼A∨B)∧A]∨A
Using the Distributive Law:
∼[(∼A∧A)∨(B∧A)]∨A
Since ∼A∧A is always False:
∼[F∨(B∧A)]∨A
∼(B∧A)∨A
Using De Morgan's Law:
(∼B∨∼A)∨A