GATE CSE 1989 | Question: 14a

Question

Symbolize the expression "Every mother loves her children" in predicate logic.

Comments

if we define (a,b) ,if a loves b.Then   

for all  M st (M,C) ?? Is it correct?

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∀x(M(x)→∀y(C(y,x)→L(x,y))) or ∀x∀y(M(x)→(C(y,x)→L(x,y))) ARE EQUAVELENT

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https://youtu.be/s41PLmHGf08?list=PLIPZ2_p3RNHidhFvyKH4U_VE1rFAyvG8U&t=2296

Answer 1

$M(x)\rightarrow x\text{ is mother}$
$Ch(y,x)\rightarrow y\text{ is child of } x$
$Lo(x,y)\rightarrow x\text{ loves }y$

$$ \forall x\forall y\ [(M(x)\wedge Ch(y,x))\rightarrow Lo(x,y)]$$

Comments

yeah that's correct

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Domain: all mothers

$love\_her\_childeren(x) : x$ love her chillderens

$\forall_x \\\ love\_her\_childeren(x)$

Will i get marks for this  :)

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@Negan no you are wrong in 2nd part :- ∀x(M(x)⟹∀y(C(y,x)∧L(x,y)))

Your expression in English will be translated as: for all x in the domain, if x is a mother, then for all y in the domain, all y's are children of x and x loves y.

When you use the AND operator with a universal quantifier, the predicate inside should be true for every possible object in the domain. But here all y need not be children of x. In the question it is stated that every mother loves her children, so x should be related to only those y who are her children.

Correct statement would be ∀x(M(x)→∀y(C(y,x)→L(x,y))) OR ∀x∀y(M(x)→(C(y,x)→L(x,y))).

They both are logically equivalent, since y doesn't matter in M(x).

Answer 2

I think this is the approach for this question.

the first statement is the correct answer for this question.

$\text{"Every mother loves her children.}$"

Rewrite The statement as,

$\text{"For every lady in this world, if a lady is a mother then she loves her children."}$

Now we introduce variable $x$,

$\text{"For every lady x in This world, if lady x is a mother then x loves her children."}$

Domain: All ladies in this world.

$L(x)\to \text{ x loves her children.}$

$M(x)\to \text{x is mother.}$

$\forall _{x} \big(M(x)\to L(x)\big)$

We use,  $\forall_{x}\big(M(x)\to L(x)\big)$  then it means that

$\text{"All ladies are mother and loves their children".}$

Comments

The first statement must be true. Consider an sql table where there are some mothers that dont love their children and some mothers that love their children.

So, 2nd statement will test if 'x' is mother and will also test if 'x' loves her children seperately.Only then it will output such mothers that love their children.

But, 1st statement will just check if 'x' is mother and if yes, will output 'x'. This is what we want from the question.

So, $\nu$x [ Mother(x) -> Loves(x) ] should be right.

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We use,  ∀x(M(x)→L(x)) then it means that

"All ladies are mother and loves their children".

Does it?

I thought it means for all x in the assumed domain(every people or every woman), if x is a mother then she loves her child.

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This would help you in getting clarity about use of → ,^ along with ∀or ∃.

L(x)→ x loves her children.

M(x)→x is mother.

domain=all ladies in the world..

∀x(M(x)→L(x)) stands for “If x is mother,then she loves her childen.

∀x(M(x)^L(x)) stands for “Every lady is a mother and loves her childen.

∃x(M(x)->L(x) stands for If the set contains ladies who are mothers,among those mothers there is atleast a mother who  loves her children.(Note that the logic statement holds true is set contains no mothers also.) 

∃x(M(x)^L(x)) stands for There is atleast one lady such that she is a mother and she loves her children.

To understand when to use →  or ^,first look at the domain ,then see what level of restriction u need to impose on logical statements beside the connective,,If a^b is used,both statements a,b must hold true,,a→ b,this relaxes the restriction on ‘a’.since if a is false,automatically a->b turns out true..

IF the whole domain needs to satisfy the property represented by ‘a’,we often use ∀x(a(x)^…..

If some x satisfies property a(x),then b(x) must hold true at any cost..for these type of statements we use ∀x(a(x)→b(x))..

SImilarly you can think for existential also.

most often universal quantifier goes with → and existential goes with ^...

 

 

 

 

Answer 3

In this question, answer can vary based on the universe of discourse.

$(i)$ When UOD contains all mothers

$M(x):$ $x$ is a mother, $L(x):$ $x$ loves their children

$\equiv\{M(1)\rightarrow L(1)\} \land$ $\{M(2)\rightarrow L(2)\}\land $ $\{M(3)\rightarrow L(3)\}...$

$\equiv\forall x\{M(x)\rightarrow L(x)\}$

 

$(ii)$ When UOD contains all people

$M(x)\rightarrow $ x is a mother
$L(x,y)\rightarrow$ x loves y
$Child(x,y)\rightarrow$ x is a child of y

Same mother can have one or more children so fix mother and vary children

$\equiv \{(M(x)\land Child(1,x))\rightarrow L(x,1)\}$ $\land\{(M(x)\land Child(2,x))\rightarrow L(x,1)\}...$

$\equiv \forall y\{(M(x)\land Child(y,x))\rightarrow L(x,y)\}$

And for every mother,

$\equiv \forall x\forall y\{(M(x)\land Child(y,x))\rightarrow L(x,y)\}$

Comments

exactly what i am thinking ..thanks bro

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In your solution, it is also possible that x is mother and y is children but they are still not related and mother can love other children also as it is not specified in your solution that x is the mother of y.

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@vizzard110 thanks for correcting me

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beautiful

Answer 4

$x \in women, y\in children$

$M(x,y) : x \hspace{1mm} is \hspace{1mm}y's \hspace{1mm} mother$

$L(x,y) : x \hspace{1mm} loves \hspace{1mm} y$

So, “Every mother loves her children” can be expressed as:

$\forall x\forall y(M(x,y)\rightarrow L(x,y))$

Comments

Right. Based on the predicates you have taken, it is correct answer.

Answer 5

Predicates: Mother(x):X is a mother
child(y,x)=Y is child of X
love(x,y)=X loves Y
Thought process
For all x (mother (x) → x loves all of her child ) 
=For all x (mother (x) → All  child of x will be loved by x)
=For all x(mother (x)→ for all y (child(y,x)→love(x,y)) [ANSWER]

Answer 6

Step 1: Identify key entities

              We are talking about mothers and children.

              Relation between them is "x is the mother of y".

              Action is "x loves y".

 

Step 2: Define predicates Let's define:

             Mother(x): "x is a mother."

             Child(y, x): "y is a child of x."

             Loves(x, y): "x loves y."

 

Step 3: Translate the statement

            "Every mother loves her children" means:

             For every x, if x is a mother, then for every y, if y is a child of x, then x loves y.

 

Step 4: Write in predicate logic

             ∀x (Mother(x) → ∀y (Child(y, x) → Loves(x, y)))

 

Step 5: Intuition

             Mother(x) ensures we only consider mothers.

             Child(y, x) ensures we only consider x's children.

             Loves(x, y) says that the mother loves her child.

 

Final Symbolization:✅

∀x (Mother(x) →∀y (Child(y, x) → Loves(x, y)))