**Example1:**

For a person p, let w(p), A(p,y), L(p) and J(p) denote that p is a woman, p admires y, p is a lawyer and p is a judge respectively. Which of the following is the correct translation in first order logic of the sentence: "All woman who are lawyers admire some judge"?

A) ∀x:[(w(x)ΛL(x))⇒(∃y:(J(y)Λw(y)ΛA(x,y)))]

B) ∀x:[(w(x)⇒L(x))⇒(∃y:(J(y)ΛA(x,y)))]

C) ∀x∀y:[(w(x)ΛL(x))⇒(J(y)ΛA(x,y))]

D) ∃y∀x:[(w(x)ΛL(x))⇒(J(y)ΛA(x,y))]

E) ∀x:[(w(x)ΛL(x))⇒(∃y:(J(y)ΛA(x,y)))]

**Solution:**

Just translating to English:

A) Every women who is a lawyer admires some women judge.

B) If a person being women implies she is a lawyer then she admires some judge. OR If a person is not women or is a lawyer he/she admires some judge.

C) Every women who is a lawyer admires every judge.

D) There is some judge who is admired by every women lawyer.

E) Every women lawyer admire some judge.

So, option (E) is the answer.

**Example 2:**

Which of the following is NOT necessarily true? { Notation: The symbol ''¬''notes negation; P(x,y) means that for given xx and yy, the property P(x,y) is true }.

A) (∀x∀yP(x,y))⇒(∀y∀xP(x,y))

B) (∀x∃y¬P(x,y))⇒¬(∃x∀yP(x,y))

C) (∃x∃yP(x,y))⇒(∃y∃xP(x,y))

D) (∃x∀yP(x,y))⇒(∀y∃xP(x,y))

E) (∀x∃yP(x,y))⇒(∃y∀xP(x,y))

**Solution:**

Option E is not necessarily true.

**Example 3:**

Consider the first-order logic sentence F:∀x(∃yR(x,y)). Assuming non-empty logical domains, which of the sentences below are implied by FF?

∃y(∃xR(x,y))

∃y(∀xR(x,y))

∀y(∃xR(x,y))

¬∃x(∀y¬R(x,y))

A) IV only

B) I and IV only

C) II only

D) II and III only

**Solution:**

option B is true

1st Method: F:∀x(∃yR(x,y))

Take option 4: ¬∃x(∀y¬R(x,y))

≡∀x(∃yR(x,y)) ((Since we know that ¬∀x≡∃x And ¬∃x=∀x)

**Example 4:**

Which one of the following well-formed formulae in predicate calculus is NOT valid ?

A) (∀xp(x)⟹∀xq(x))⟹(∃x¬p(x)∨∀xq(x))

B) (∃xp(x)∨∃xq(x))⟹∃x(p(x)∨q(x))

C) ∃x(p(x)∧q(x))⟹(∃xp(x)∧∃xq(x))

D) ∀x(p(x)∨q(x))⟹(∀xp(x)∨∀xq(x))

**Solution:**

Here, (D) is not valid

Let me prove by an example

What (D) is saying here is:

For all x ( x is even no or x is odd no ) ⟹ For all x( x is even no ) or For all x ( x is odd no)

OR

If every x is either even or odd, then every x must be even or every x must be odd.

If our domain is the set of natural numbers LHS is true but RHS is false as not all natural numbers are even or odd.

Check the last few examples on Nested qualifiers

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