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29 votes
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Symbolize the expression "Every mother loves her children" in predicate logic.
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4 Answers

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

$$ \forall x\forall y\ [(M(x)\wedge C(y,x))\rightarrow Lo(x,y)]$$
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22 votes
22 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

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

edited by
14 votes
14 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,

$\equiv \forall x\forall y\{(M(x)\land Child(y,x))\rightarrow L(x,y)\}$
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4 votes
4 votes
$x \in women, y\in children$

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

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

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

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

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