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Excuse me, who has the final examination paper of discrete and analog circuits in Nantong University Computer College this year?
Discrete mathematics simulation test paper one answer

Short answer questions (25%):

Solution: The relation matrix of R is expressed as:

Reflexive closure: r (r) = {

Symmetric closure s (r) =

Transitive closure t (r) =

Solution:

⑴(P∨? P)→? The truth table of q is as follows:

PQP∨? P(P∨? P)→? Q

0 0 1 1

0 1 1 0

1 0 1 1

1 1 1 0

Is a satisfiable formula.

⑵(P→Q)(? Q→? The truth table of p) is as follows:

PQ(P→Q)(? Q→? P)(P→Q)(? Q→? p)

0 0 1 1 1

0 1 1 1 1

1 0 0 0 1

1 1 1 1 1

This is an eternal formula.

⑶G=(P∨Q)∧(? P∨Q)∧(P∨? Q)∧(? P∨? Q) The truth table is as follows:

PQ(P∨Q)(? P∨Q)(P∨? Q) (? P∨? q)G

0 0 0 1 1 1 0

0 1 1 1 0 1 0

1 0 1 0 1 1 0

1 1 1 1 1 0 0

This is a formula that is always wrong.

⑷G=(P∧Q)∨(? P∧Q)∨(P∧? Q)∨(? P∧? Q) The truth table is as follows:

PQ(P∧Q)(? P∧Q)(P∧? Q) (? P∧? q)G

0 0 0 0 0 1 1

0 1 0 1 0 0 1

1 0 0 0 1 0 1

1 1 1 0 0 0 1

This is an eternal formula.

Solutions;

It's not a lattice. There are two elements that have no lower bounds.

Is a lattice, and any two elements have a minimum upper bound and a maximum lower bound.

This is the same reason as (b).

It's not a lattice. There are two elements that have no lower bounds.

True or false (25%):

1. solution:

The (1) proposition is false. R∩S is reflexive, but ROS is not necessarily reflexive. For example, r = {

(2) The proposition is false. R∩S is transitive, but ROS is not necessarily transitive. For example: set A={ 1, 2, 3, 4, 5}, r = {

(3) The proposition is true.

(4) The proposition is false. For example, let A={a, b, c} and r = {

2. solution: because for any x, y∈R, |x-y|=|y-x|, so x*y=y*x, so * is interchangeable. For 1, 2,3 ∈ r, (1* 2) * 3 = ||1-2 |-3 | = 2, while1* (2 * 3) = |/kloc-. Because for every x∈R, it is impossible to make x*y=y hold for any y∈R, so there is no unitary about * in R.

3. Which of the following pictures have Euler paths?

Euler path

Euler path

There is no Euler path.

There is no Euler path.

Proof problem (35%):

Prove:

∵fOg=IA is bijective, ∴f is injective, g is surjective,

∵gOf=IB is bijective, ∴g is injective and F is surjective.

So g and f are bijective, so g and f are reversible.

gOfOf- 1 =(gOf Of- 1 = IBOf- 1 = f- 1

Gofof-1= go (fof-1) = goia = g, so g=f- 1.

Foog- 1 = (fog) og-1= iaog-1= g-1

Foog-1= f o (gog-1) = foib = f, so f = g- 1.

Prove:

P∨Q P

p→QT,①,E

q→SP

p→ST,②,③,I

s→PT,④,E

p→RP

s→RT,⑤,⑥,I

SVRT,⑦,E

Prove:

Q(x)P

(x)(? P(x))P

(x)(? q(x))T,①,E

(x)(? P(x))∧(x)(? q(x))T,②,③,I

(x)(? P(x)∧? q(x))T,④,E

P(y)∧? Ask (y) us, ⑤

p(y)∨Q(y))T,⑥,E

(x)(P(x)∨Q(x))P

p(y)∞Q(y)US,⑧

(P(y)∨Q(y))∧? (P(y)∨Q(y))T,⑦,⑨,I

Prove:

known

For any a, b∈G, there is (a * b) * (a * b) = e.

a*(b*(a*b))=a*a

b*(a*b)=a

Then b * (b * (a * b)) = b * a a.

e*(a*b)=b*a

a*b=b*a

So as to satisfy the exchange law.

So < g; *> This is an Abelian group.

Comprehensive questions (15%):

Solution:

⑴ P→(P∧(Q→P)) P→(P∧(? Q∨P))

? P∨(P∧(P∨? Q)

? P∨P 1

∴ p→ (p ∧ (the main conjunctive normal form of Q→ p)) is empty.

The main disjunctive paradigm is (? P∧Q)∨(P∧? Q)∨(P∧Q)∨(? P∧? Q)

⑵ (Q→P)∧(? P∧Q)(? Q∨P)∧(? P∧Q)

(? Q∨P)∧? (? Q∨P) 0

∴(Q→P)∧(? The principal disjunctive normal form of P∧Q) is empty.

Lord conjunctive normal form is (? P∨Q)∧(P∨? Q)∧(P∨Q)∧(? P∨? Q)

Solution: The picture below shows what may happen in the game. In the picture, those marked with A represent the first victory, and those marked with B represent the second victory. This is a complete binary tree.