QS solve - Mahbub Sir

নিচের প্রায় সবগুলো উত্তরই LLM generated, তবে যদি কোনোটা নিয়ে সংশয় সৃষ্টি হয়, feel free to ask!

Equivalence Check

একটা কথা মাথায় রাখতে হবে যে $\models$ এই চিহ্নটা হলো logical entails, বাম পাশে সত্য আর ডান পাশে মিথ্যা এরকম কখনোই হতে পারবে না ;)

আর == এটা হলো equivalence যেটা আমরা tpointtech এ পড়েছি

i. $(A\wedge B)\models (A\iff B)$

This statement is correct.

• Clarification: The symbol $\models$ means “logically entails.” This statement asserts that in every situation where $(A\wedge B)$ is true, $(A\iff B)$ must also be true.

  • For $(A\wedge B)$ to be true, both A and B must be true.

  • The expression $(A\iff B)$ is true if and only if A and B have the same truth value.

  • Therefore, in the specific case where both A and B are true, $(A\iff B)$ is also true. This confirms the entailment.

  We can also see this with a truth table:

A	B	A∧B	A⇔B
T	T	T	T
T	F	F	F
F	T	F	F
F	F	F	T

The only row where $(A\wedge B)$ is true is the first row, and in that row, $(A\iff B)$ is also true.

---

ii. $A\iff B\models \neg A\vee B$

This statement is correct.

• Clarification: We need to check if $\neg A\vee B$ is true in all cases where $A\iff B$ is true.

  • $A\iff B$ is true under two conditions:

    1. Both A and B are true.

    2. Both A and B are false.

  • Let’s check the expression $\neg A\vee B$ (which is equivalent to $A\Rightarrow B$) for these two conditions:

    1. If A=True and B=True, then $\neg T\vee T\Rightarrow F\vee T$, which is True.

    2. If A=False and B=False, then $\neg F\vee F\Rightarrow T\vee F$, which is True.

  • Since $\neg A\vee B$ is true in all cases where $A\iff B$ is true, the entailment holds.

---

iii. $(A\vee B)\wedge (\neg C\vee \neg D\vee E)\models (A\vee B\vee C)\wedge (B\wedge C\wedge D\Rightarrow E)$

This statement is correct.

• Clarification: Let the left-hand side (LHS) be our premises and the right-hand side (RHS) be our conclusion. The LHS gives us two premises:

  1. $P_1:(A\vee B)$

  2. $P_2:(\neg C\vee \neg D\vee E)$

  The RHS is a conjunction, so we must be able to derive both parts from the premises.

  • Part 1 of RHS: $(A\vee B\vee C)$

    • We can derive this directly from $P_1$ using the Rule of Addition (or Disjunction Introduction), which states that if we have $P$, we can infer $P\vee Q$. Since we have $(A\vee B)$, we can validly infer $(A\vee B)\vee C$.

  • Part 2 of RHS: $(B\wedge C\wedge D\Rightarrow E)$

    • First, convert the implication to its equivalent disjunctive form: $\neg (B\wedge C\wedge D)\vee E$, which simplifies to $(\neg B\vee \neg C\vee \neg D\vee E)$ using De Morgan’s laws.

    • We can derive this from $P_2$ using the Rule of Addition. Since we have $(\neg C\vee \neg D\vee E)$, we can validly infer $\neg B\vee (\neg C\vee \neg D\vee E)$.

  Since both parts of the RHS conclusion can be logically derived from the LHS premises, the entire statement is correct.

---

iv. $(A\wedge B)\Rightarrow C\models (A\Rightarrow C)\vee (B\Rightarrow C)$

This statement is correct.

• Clarification: This is a case of logical equivalence. If two expressions are logically equivalent, they also entail each other. We can prove this by converting both sides into a standard form.

  • Left-Hand Side (LHS): $(A\wedge B)\Rightarrow C$

    • $\equiv \neg (A\wedge B)\vee C$ (Implication Equivalence)

    • $\equiv (\neg A\vee \neg B)\vee C$ (De Morgan’s Law)

    • $\equiv \neg A\vee \neg B\vee C$

  • Right-Hand Side (RHS): $(A\Rightarrow C)\vee (B\Rightarrow C)$

    • $\equiv (\neg A\vee C)\vee (\neg B\vee C)$ (Implication Equivalence)

    • $\equiv \neg A\vee C\vee \neg B\vee C$

    • $\equiv \neg A\vee \neg B\vee C$ (by Idempotent Law, where $C\vee C\equiv C$)

  Since both the LHS and RHS simplify to the exact same logical expression $(\neg A\vee \neg B\vee C)$, they are logically equivalent, and the entailment is correct.

---

v. $A\iff B\models A\vee B$

❌ Incorrect.

• Reasoning: The entailment fails when A and B are both false.

  • If A=False and B=False, then $(A\iff B)$ is true.

  • However, $(A\vee B)$ becomes $(F\vee F)$, which is false.

  • Since we found a case where the premise is true but the conclusion is false, the statement is incorrect.

---

vi. $A\Rightarrow B\models \neg A\vee B$

✅ Correct.

• Reasoning: This is a fundamental logical equivalence. The expression $A\Rightarrow B$ (A implies B) is defined as being logically equivalent to $\neg A\vee B$. Since they are equivalent, they entail each other.

---

iv. $(A\vee B)\wedge (\neg C\vee \neg D\vee E)\models (A\vee B\vee C)\wedge (B\wedge C\wedge D\Rightarrow E)$

✅ Correct.

• Reasoning: Both parts of the conclusion on the right side can be logically derived from the premises on the left side using the Rule of Addition, which allows you to add any proposition to an existing one with an ‘OR’ operator. For example, since we know $(A\vee B)$ is true, we can infer that $(A\vee B\vee C)$ must also be true.

---

v. $(A\vee B)\wedge (\neg C\vee \neg D\vee E)\models (A\vee B)\wedge (\neg D\vee E)$

❌ Incorrect.

• Reasoning: This statement claims that if $(\neg C\vee \neg D\vee E)$ is true, then $(\neg D\vee E)$ must also be true. This is not valid.

  • Counterexample: Let C=False, D=True, and E=False.

  • The premise $(\neg C\vee \neg D\vee E)$ becomes $(T\vee F\vee F)$, which is true.

  • However, the conclusion $(\neg D\vee E)$ becomes $(F\vee F)$, which is false.

---

vi. $(A\vee B)\wedge \neg (A\Rightarrow B)$ is satisfiable

✅ Correct.

• Reasoning: An expression is satisfiable if there’s at least one assignment of truth values that makes it true. Let’s simplify the expression:

  • $(A\vee B)\wedge \neg (\neg A\vee B)$

  • $(A\vee B)\wedge (A\wedge \neg B)$ (Using De Morgan’s Law)

• For this entire expression to be true, the part $(A\wedge \neg B)$ must be true, which means A must be True and B must be False.

• This assignment (A=T, B=F) makes the entire expression true, so it is satisfiable.

---

vii. $(A\wedge B)\Rightarrow C\models (A\Rightarrow C)\vee (B\Rightarrow C)$

✅ Correct.

• Reasoning: The expressions on the left and right are logically equivalent. Both can be simplified to the same form: $\neg A\vee \neg B\vee C$. Since they are equivalent, the entailment is correct.

First Order Logic

A. Every cat loves its mother or father.

• (i) ∀x Cat(x) ⇒ Loves(x, Mother(x) v Father(x))

  • (2) Syntactically invalid. The arguments to a predicate like Loves must be terms (variables, constants, or functions). The expression Mother(x) v Father(x) is a logical disjunction, not a term. You cannot “love” a logical “or”.

• (ii) ∀x ¬Cat(x) v Loves(x, Mother(x)) v Loves(x, Father(x))

  • (1) Correctly expresses the sentence. This is logically equivalent to the standard form ∀x Cat(x) ⇒ (Loves(x, Mother(x)) v Loves(x, Father(x))). The equivalence rule used here is that P ⇒ Q is the same as ¬P v Q.

• (iii) ∀x Cat(x) ∧ (Loves(x, Mother(x)) v Loves(x, Father(x)))

  • This expression is syntactically valid, but it does not correctly express the sentence. It claims that everything in the universe is a cat (∀x Cat(x)), which is incorrect. The proper logical connective to use with a universal quantifier (∀) for “Every…” statements is implication (⇒), not conjunction (∧).

---

B. Every dog who loves one of its brothers is happy.

• (i) ∀x Dog(x) ∧ (∃y Brother(y, x) ∧ Loves(x, y)) ⇒ Happy(x)

  • (2) Syntactically invalid. The scope of the universal quantifier ∀x is not correctly applied to the entire sentence. As written, the x in Happy(x) is a “free variable” because it’s outside the scope of ∀x. The entire expression should be enclosed in brackets after the ∀x.

• (ii) ∀x [Dog(x) ∧ (∃y Brother(y, x) ∧ Loves(x, y))] ⇒ Happy(x)

  • (1) Correctly expresses the sentence. This translates to: “For any x, if x is a dog AND there exists a y such that y is a brother of x AND x loves y, THEN x is happy.” This perfectly captures the sentence’s meaning.

• (iii) ∀x Dog(x) ∧ [∀y Brother(y, x) ⇒ Loves(x, y)] ⇒ Happy(x)

  • (2) Syntactically invalid. This has the same scope issue as expression (i), where the x in Happy(x) is a free variable. Additionally, it incorrectly uses a universal quantifier (∀y), which would mean the dog loves all of its brothers, not just one of them.

---

C. No dog bites a child of its owner.

• (i) ∀x Dog(x) ⇒ ¬Bites(x, Child(Owner(x)))

  • (2) Syntactically invalid. The expression Child(Owner(x)) is being used as a function that returns a single object (“the child”), but the original sentence implies any child. Logic requires a predicate Child(y, z) (y is a child of z) and a quantifier for the child (y).

• (ii) ¬∃x, y Dog(x) ∧ Child(y, Owner(x)) ∧ Bites(x, y)

  • (1) Correctly expresses the sentence. This reads: “There does not exist an x and a y such that x is a dog, y is a child of x’s owner, and x bites y.” This is an accurate way to state the original sentence.

• (iii) ∀x Dog(x) ⇒ (∀y Child(y, Owner(x)) ⇒ ¬Bites(x, y))

  • (1) Correctly expresses the sentence. This is a logically equivalent and correct formulation. It reads: “For any x, if x is a dog, then for any y, if y is a child of x’s owner, then x does not bite y.”

• (iv) ¬∃x Dog(x) ⇒ (∃y Child(y, Owner(x)) ∧ Bites(x, y))

  • (2) Syntactically invalid. The scope of ¬∃x only covers Dog(x). The variables x that appear later in the expression (Owner(x) and Bites(x, y)) are free and not bound by any quantifier.

---

D. Everyone’s zip code within a state has the same first digit.

• (i) ∀x, s, z1 [State(s) ∧ LivesIn(x, s) ∧ Zip(x)=z1] ⇒ [∀y, z2 LivesIn(y, s) ∧ Zip(y)=z2 ⇒ Digit(1, z1)=Digit(1, z2)]

  • (1) Correctly expresses the sentence. Although it’s written in a complex way by introducing z1 and z2 to represent the zip codes, it is logically sound. It states that for any person x in a state s with zip z1, any other person y in the same state with zip z2 will have the same first digit.

• (ii) ∀x, s [State(s) ∧ LivesIn(x, s) ∧ ∃z1 Zip(x)=z1] ⇒ [∀y, z2 LivesIn(y, s) ∧ Zip(y)=z2 ∧ Digit(1, z1)=Digit(1, z2)]

  • (2) Syntactically invalid. The variable z1 is introduced with an existential quantifier (∃z1) in the antecedent (the if part), but it is then used in the consequent (the then part). A variable’s scope does not extend outside of the clause where its quantifier is located.

• (iii) ∀x, y, s State(s) ∧ LivesIn(x, s) ∧ LivesIn(y, s) ⇒ Digit(1, Zip(x)=Zip(y))

  • (2) Syntactically invalid. The function Digit(...) expects a zip code as its second argument. The expression Zip(x)=Zip(y) is a logical comparison that evaluates to true or false; it is not a zip code and cannot be an argument for the Digit function.

• (iv) ∀x, y, s (State(s) ∧ LivesIn(x, s) ∧ LivesIn(y, s)) ⇒ Digit(1, Zip(x))=Digit(1, Zip(y))

  • (1) Correctly expresses the sentence. This is the most clear and direct translation. It says: “For any two people x and y and any state s, if s is a state and both x and y live in s, then the first digit of x’s zip code is equal to the first digit of y’s zip code.”

First order Logic - Translation

📝 Predicate Definitions

প্রশ্ন উত্তরে সুবিধার্থে কিছু function ব্যবহার করা যেতে পারে, এটা optional, যার যেভাবে খুশি ধরে নেয়া যাবে…

• Gardener(x): $x$ is a gardener.

• Likes(x, y): $x$ likes $y$.

• TheSun: A constant representing the sun.

• Person(x): $x$ is a person.

• Time(t): $t$ is a time.

• CanFool(x, t): Person $x$ can be fooled at time $t$.

• Mushroom(x): $x$ is a mushroom.

• Purple(x): $x$ is purple.

• Poisonous(x): $x$ is poisonous.

• Clinton: A constant representing Clinton.

• Tall(x): $x$ is tall.

• Man(x): $x$ is a man.

• Fool(x): $x$ is a fool.

• Surgeon(x): $x$ is a surgeon.

• Lawyer(x): $x$ is a lawyer.

• Actor(x): $x$ is an actor.

• Doctor(x): $x$ is a doctor.

• HoldsAnotherJob(x): $x$ holds a job in addition to one already mentioned.

• BossOf(x, y): $x$ is the boss of $y$.

• CustomerOf(x, y): $x$ is a customer of $y$.

---

Set - I

i. Every gardener likes the sun.

• Answer: $\forall x(Gardener(x)\to Likes(x,TheSun))$

ii. You can fool some of the people all of the time.

• Answer: $\exists x(Person(x)\wedge \forall t(Time(t)\to CanFool(x,t)))$

iii. You can fool all of the people some of the time.

• Answer: $\forall x(Person(x)\to \exists t(Time(t)\wedge CanFool(x,t)))$

iv. All purple mushrooms are poisonous.

• Answer: $\forall x((Mushroom(x)\wedge Purple(x))\to Poisonous(x))$

v. No purple mushroom is poisonous.

• Answer: $\forall x((Mushroom(x)\wedge Purple(x))\to \neg Poisonous(x))$

vi. There are exactly two purple mushrooms.

• Answer: $\exists x\exists y((Mushroom(x)\wedge Purple(x))\wedge (Mushroom(y)\wedge Purple(y))\wedge (x\ne y)\wedge \forall z((Mushroom(z)\wedge Purple(z))\to (z=x\vee z=y)))$

vii. Clinton is not tall.

• Answer: $\neg Tall(Clinton)$

viii. All men are not fool.

• Answer: $\exists x(Man(x)\wedge \neg Fool(x))$

  Note: This translation assumes the sentence means “Not all men are fools.” The English is slightly ambiguous. If it were to mean “No man is a fool,” the translation would be $\forall x(Man(x)\to \neg Fool(x))$.

---

Set - II

i. Emily is either a surgeon or a lawyer.

• Answer: Surgeon(Emily) ∨ Lawyer(Emily)

ii. Joe is an actor, but he also holds another job.

• Answer: Actor(Joe) ∧ HoldsAnotherJob(Joe)

  Note: “But” is translated as a logical AND (∧). The concept of “another job” is represented with a dedicated predicate for simplicity.

iii. All surgeons are doctors.

• Answer: $\forall x(Surgeon(x)\to Doctor(x))$

iv. Joe does not have a lawyer (i.e., is not a customer of any lawyer).

• Answer: $\neg \exists x(Lawyer(x)\wedge CustomerOf(Joe,x))$

  • This can also be written equivalently as: $\forall x(Lawyer(x)\to \neg CustomerOf(Joe,x))$

v. Emily has a boss who is a lawyer.

• Answer: $\exists x(BossOf(x,Emily)\wedge Lawyer(x))$

vi. There exists a lawyer all of whose customers are doctors.

• Answer: $\exists x(Lawyer(x)\wedge \forall y(CustomerOf(y,x)\to Doctor(y)))$

vii. Every surgeon has a lawyer.

• Answer: $\forall x(Surgeon(x)\to \exists y(Lawyer(y)\wedge CustomerOf(x,y)))$