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## Here's the Solution to this Question

a)

U- set of allpeople

C- students of this class

P:U->{0,1}- predicat,P(X)=1~man X$\in$ U has Honda bike.

Q:U->{0,1}-predicat,Q(X)=1~ man X has gotten at least one traffic violation challen.

Given that $P(Asim),\forall(X\in U)(P(X)->Q(X))$

From specialization rule we have $P(Asim),P(Asim)->Q(Asim)$

Therefore by modus ponens rule Q(Asim), but $Asim\in C$

By definition $\exist$ we have $\exist (X\in C) Q(X).$

b)

U- set of all students.

P:U->{0,1}-predicat, P(X)~X has taken discret structure course

Q:U->{0,1}-predicat ,Q(X)~ X can take algorithm course.

C- set of five roommates, |C|=5,

Statement is

$P(Azhar),P(Haris),P(Nouman), \forall(X\in U)(P(X)->Q(X))->\forall(Y\in C)Q(Y)$

is not true because there is such interpretation that left part true but right part is false. Evidently C\{Azhar,Haris,Nouman}$\ne\empty$ .

Let $a\in C$ \{Azhar,Haris,Nouman},in interpretation we assume P(a)=Q(a)=0,

P(Azhar)=P(Haris)=P(Nouman)=Q(Azhar)Q=(Haris)=Q(Nouman)=1

Then $\forall(Y\in C)Q(Y)=0$ and $P(a)->Q(a)=\overline{P(a)}\lor Q(a)=\overline 0 \lor 1=1$ , therefore

$P(Azhar),P(Haris),P(Nouman), \forall(X\in U)(P(X)->Q(X))->\forall(Y\in C)Q(Y)=1$

Thus statement is not true.

c)

P(X)- movie X is wonderful.

Q(X) movie X produced by John Sayles.

R(X) movie X is about coal miners.

U-set of all movies.

Statement is:

$\forall(X\in U)(Q(X)\rarr P(X)),\space \exist(X\in U)(Q(X)\land R(X))\rarr \\ \exist(X\in U)(R(X)\land P(X))$

Proof:

$\exist(X\in U)(Q(X)\land R(X))\implies Q(a)\land R(a)$

concretization;

$\forall(X\in U)(Q(X)\rarr P(X))\implies Q(a)\rarr P(a)$

spesialization

$Q(a)\land R(a)\implies Q(a)$

conjuction rule

$Q(a),Q(a)\rarr P(a)\implies P(a)$

modus ponens

$Q(a)\land R(a)\implies R(a)$

conjuction rule

$P(a),R(a)\implies P(a)\land R(a)$

build conjuction rule

$P(a)\land R(a)\implies \exist(X\in U)(R(X)\land P(X))$

build existence rule

d)

P(X)~ X has been to Saudi Arabia

Q(X)~X performs Umrah

U- set of all people

C- this class

Statement:

$\exist (X\in C)P(X),\space \forall(X\in U)(P(X)\rarr Q(X))\implies \exist(X\in C)Q(X)$

Proof:

a$\in$ C, P(a)- conctetization'

P(a)->Q(a)- specialization;

$P(a),P(a)\rarr Q(a)\implies Q(a)$

modus ponens

$Q(a)\implies \exist(X\in C)Q(X)$

definition of $\exist$