GATE1989-14a

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?

Answer 1

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

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

Comments

∀x∀y [(M(x)∧C(y,x))→Lo(x,y)]

 

 

"∀y" why u used this before the bracket ??

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Then where you want to use it?

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actually not   " I ".....according to the question

∀x(M(x)→L(x))  , it means that

"All ladies are mother and loves their children"

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All ladies are mother and loves their children.

See carefully there is word "their".

 

∀x(M(x)→L(x)) logic given says, all those things in universe which are x are mother and love x i.e. themselves.

Firstly we need atleast two variables x and y for mother and children, and then we will establish relationships between them, which are following

1. Whatever x we select from universal set, should be a mother ¥x[M(x)].

2. Then that mother should have "her own" as in question it is written "mother love their children not all childern of the world which might include children of other mother too, a mother need to love only her childrens" so Ch(x,y) i.e. x is mother of y.

3. Mother love her children(s) Lo(x,y).

So, all x which are mother and for all children y whoes mother is x, x mother love child y.

¥x¥y runs like a two for loops, first a mother will be selected then inner loop will run for children, who so ever y satisfy Ch(x,y) i.e. choosen x in outer loop is mother of select child in inner loop, it will checked that she loves her  that child by Lo(x,y), the y++ check for next y, if he/she is mother x child and check for love condition, and loop goes on and on.

I think you'll get now why above giben is correct.

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thanks bro... for grt effort

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What if we do like

∀y∀x [(M(x)∧C(y,x))→Lo(x,y)]

will it affect ??

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@jatin khachane 1

It won't make any difference but writing "∀x∀y [(M(x)∧C(y,x))→Lo(x,y)]" would just enhance the meaning. 

This means we compare one mother will all the children available in the universe of discourse.

Let there be "n" women in the universe of discourse(a set) containing women and "m" children in another universe of discourse containing children.

Picks up a mother and checks her with all the children
for(x=1;x<n;x++) //Take each woman "x" from universe of discourse containing 
                  //women
{       
if(M[x]==0)// Check if she is a mother or not
 { flag=1
   break;
 }
for(y=1,y<m;y++)  //Take each children "y" from universe of discourse containing
                 // children    
 {
    if(C[y][x]==1) //Check "y" is a child of that mother "x"
   { 
    if(Lo[x][y]==1) //Check if "x" loves "y"
    flag=1
    else 
    flag=0;
   }
   else 
    flag=1;
 }  
}
 

This one "∀y∀x [(M(x)∧C(y,x))→Lo(x,y)] " means we compare each child with all the women available in the universe of discourse whether she is his/ her mother and whether she loves him/her or not.

Picks up a child and checks him/her with all the mother
for(y=1;y<m;y++) //Take each child "y" from universe of discourse containing 
                  //children
{       
for(x=1,x<n;x++)  //Take each woman "y" from universe of discourse containing
                 // women    
 {
     if(M[x]==0)// Check if she is a mother or not
     { flag=1
       break;
      }
    if(C[y][x]==1) //Check "y" is a child of "x"
   { 
    if(Lo[x][y]==1) //Check if "x" loves "y"
    flag=1
    else 
    flag=0;
   }
   else 
    flag=1;
 }  
}

 

The result does not change.

 

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@minipanda :/

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 bhuv Sir, What is difference between ^ and -> (AND and implies) operator. When to use which operator. Please explain.

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@anchitjindal07

Statement1 : All men are mortal.

$\forall x[Men(x) \Rightarrow Mortal(x)]$

Statement2: Some trigonometric functions are continuous.

$\exists x[Trigo(x) \wedge Continuous(x)]$

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@MiNiPanda sir, can you please explain the use of “flag variable”, that you used in the program written by you in comment.

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Let $M(x)$ denote “x is a mother”,

$C(y,x)$ denote “y is a child of x”,

$L(x,y)$ denote “x loves y”.

If the domain of discourse consists of all persons, then the given expression may also be stated as

$\forall x(M(x)\rightarrow \forall y(C(y,x)\rightarrow L(x,y)))$    [as per my understanding]

Please correct me if I’m wrong :)

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@GATE_aspirant_2021 I think you are right, 

as @abichandani said

Statement1 : All men are mortal.

∀x[Men(x)⇒Mortal(x)]  , with “∀x” we used “⇒” to restrict domain

Statement2: Some trigonometric functions are continuous.

∃x[Trigo(x)∧Continuous(x)]  , with “∃x” we used “∧” to restrict domain

 

then

M(x) denote “x is a mother”, 
C(y,x) denote “y is a child of x”,
L(x,y) denote “x loves y”.

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

should be correct because with “∀x” we use “⇒” to restrict domain and with “∃x” we use “∧” to restrict domain.

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I came up with this answer,
Please validate if it is equivalent to the Best Answer
$∀x(M(x)→∃y(C(y,x)∧L(x,y))) $

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.

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:

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