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Computer Science A67
Homework Assignment # 2
Question 1. Quantifiers [12 marks]
For each of the following logical expressions, write an English sentence with the same meaning. (Hint : Your English sentences should not mention variable names x, y, etc.) Indicate which statements are true and which are false.
For the first three expressions, the universe of discourse is all people. Let C(x,y) stand for “x is a child of y” .
1. 丫x3y, C(x,y)
2. 3x→丫y, C(y, x)
3. →3x丫y,(C(x,y) → (3z,(z ≠ x) Λ C(z, y)))
For the next three expressions, the universe of discourse is the natural numbers. Let P(x) stand for “x is prime”, and F(x,y) stand for “x is a factor of y” .
4. 3y丫x, → (y = x2 )
5. 丫x,(→P(x) → (3y,(y < x) Λ F(y, x)))
6. 丫x,(P(x) → (→3y, P(y) Λ P(x + y)))
Question 2. Negation [20 marks]
For each of the following sentences:
1. Write a logical expression that represents the English sentence.
2. Write an English sentence that is the negation of the original sentence.
3. Negate the expression in Step 1, and use logical equivalence rules to demonstrate that the result is equivalent to the logical form of the English sentence in Step 2.
Let B(x) stand for “x is a bird”, C(x) stand for “x is a cat”, D(x) stand for “x is a dog”, S(x) stand for “x is a squirrel”, A(x) stand for “x is an adult”, and L(x) stand for “x is slow” . Let E(x,y) stand for “x has eaten y” and H(x,y) stand for “x has chased y” . Let M be a variable representing my dog. The universe of discourse is animals.
1. Every dog has chased at least one squirrel.
2. At least one adult cat has eaten some birds.
3. Every squirrel has never chased any animals.
4. Slow cats have never eaten any of the birds that they chased.
5. My dog has chased squirrels but not cats.
Question 3. Deductive Reasoning [32 marks] For each of the following arguments, either:
• prove the argument is valid by using Rules of Inference, OR
• prove the argument is not valid by providing a world that makes the premises true and the conclusion false. Part (a) [8 marks]
All unicorns are either white or rainbow coloured. There is a unicorn that is not white. Therefore, there is a unicorn that is rainbow coloured.
Let U(x) stand for “x is a unicorn”, W(x) stand for “x is white”, R(x) stand for “x is rainbow coloured”, universe of discourse is creatures.
Part (b) [8 marks]
All unicorns are either white or rainbow coloured. White unicorns are magical. Rainbow unicorns can fly. Only magical creatures fly. Therefore, all unicorns are magical.
Let W(x) stand for “x is white”, R(x) stand for “x is rainbow coloured”, M(x) stands for “x is magical”, F(x) stands for “x can fly”, universe of discourse is unicorns.
Part (c) [8 marks]
Someone committed a crime. Every crime has a witness. A person can testify about a crime only if they witnessed it. If any person testifies about a crime, then the person who committed the crime goes to jail. Therefore, someone will go to jail.
Let C(x;y) be “person x commits crime y”, W(x;y) be “person x witnesses crime y”, T(x;y) be “x testifies about crime y”, J(x) be “x goes to jail” .
Part (d) [8 marks]
Everyone who witnessed a crime must testify about it. At least one person witnessed at least one crime. If any person testifies about a crime, then the person who committed the crime is found guilty. If a crime is witnessed, it was committed by someone. Everyone who is found guilty goes to jail. Therefore, someone will go to jail.
Let C(x;y) be “person x commits crime y”, W(x;y) be “person x witnesses crime y”, T(x;y) be “x testifies about crime y”, G(x) be “x is found guilty”, J(x) be “x goes to jail” .
Question 4. [32 marks]
Recall that we can prove that statements A and B are logically equivalent as follows:
1. Develop a proof of A → B: Suppose A
::: B
2. Develop of proof of B → A: Suppose B
::: A
For each of the following pairs of expressions, either prove they are equivalent by using the above technique, or prove they are not equivalent by providing a counterexample world. (Hint : you can let the universe of discourse be the set of natural numbers, and provide an example of properties P(x), Q(x) for which one expression is True and the other is False.)
1. (∀x;P(x)) ∧ (∀x;Q(x)) and ∀x;P(x) ∧ Q(x)
2. (∃x;P(x)) ∧ (∃x;Q(x)) and ∃x;P(x) ∧ Q(x)
3. (∀x;P(x)) ∨ (∀x;Q(x)) and ∀x;P(x) ∨ Q(x)
4. (∃x;P(x)) ∨ (∃x;Q(x)) and ∃x;P(x) ∨ Q(x)
5. (∀x;P(x)) → (∀x;Q(x)) and ∀x;P(x) → Q(x)
6. (∃x ∈ A;P(x)) ∧ (∃x ∈ B;P(x)) and ∃x ∈ A ∩ B;P(x)
7. (∃x ∈ A;P(x)) ∨ (∃x ∈ B;P(x)) and ∃x ∈ A ∪ B;P(x)