for all M st (M,C) ?? Is it correct?

18 votes

24 votes

Best answer

0

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

∀x(M(x)→L(x)) , it means that

"All ladies are mother and loves their children"

∀x(M(x)→L(x)) , it means that

"All ladies are mother and loves their children"

9

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.

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.

7

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.

0

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

0

Statement1 : All men are mortal.

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

Statement2: Some trigonometric functions are continuous.

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

0

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

1

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 :)

$C(y,x)$ 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 :)

0

@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 domainStatement2: 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.

14 votes

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

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

0

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.

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.

9 votes

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,

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

$(i)$ 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,

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