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[mathematical logic] normal form (conjunctive normal form | disjunctive normal form | major item | minor item | maximal item | minor item | principal conjunctive normal form | principal disjunctive no

2022-07-03 03:13:00 【Programmer community】

List of articles

One . Relevant concepts
- 1. Simple Disjunction Syntaxis type
- - ( 1 ) Simple conjunction
  - ( 2 ) Simple disjunctive
- 2. minterm
- - ( 1 ) minterm brief introduction
  - ( 2 ) minterm explain
  - ( 3 ) Two propositional variables Of minterm
  - ( 4 ) Three propositional variables Of minterm
  - ( 5 ) minterm True assignment The formula name Between Of conversion And Deduce
- 3. The maximal term
- - ( 1 ) The maximal term brief introduction
  - ( 2 ) The maximal term explain
  - ( 3 ) The maximal term of two propositional variables
  - ( 4 ) The maximal term of three propositional variables
  - ( 5 ) The maximal term False assignment The formula name Between Of conversion And Deduce
Two . title
- 1. Use equivalent calculus to find Principal disjunctive normal form and The main conjunctive paradigm
- 2. Use Truth table method seek Principal disjunctive normal form and The main conjunctive paradigm

One . Relevant concepts

1. Simple Disjunction Syntaxis type

( 1 ) Simple conjunction

Simple conjunction :

1. form : Propositional argument ( $p p ¬ p \lnot p ) ;$
$p$ ) or Propositional argument negatives (
2. Concept : A limited number Propositional argument Or its Negative form The conjunctive form of composition , be called Simple conjunction ;
3. Example :
- ① Single propositional argument :
  $p$ ;
- ② Single propositional argument negatives : $\lnot p$
  $\neg p$
- ③ Two Propositional argument Or its negative form The conjunctive form of composition : $\land \lnot q$
  $p \land \neg q$
- ④ Three Propositional argument Or its negative form The conjunctive form of composition : $\land q \land r$
  $p \land q \land r$

( 2 ) Simple disjunctive

Simple disjunctive :

1. form : Propositional argument ( $p p ¬ p \lnot p ) ;$
$p$ ) or Propositional argument negatives (
2. Concept : A limited number Propositional argument Or its Negative form Disjunctive of composition , be called Simple disjunctive ;
3. Example :
- ① Single propositional argument :
  $p$ ;
- ② Single propositional argument negatives : $\lnot p$
  $\neg p$
- ③ Two Propositional argument Or its negative form Disjunctive form : $\lor \lnot q$
  $p \lor \neg q$
- ④ Three Propositional argument Or its negative form Disjunctive form : $\lor q \lor r$
  $p \lor q \lor r$

2. minterm

( 1 ) minterm brief introduction

minterm : minterm yes A kind of Simple conjunction ;

1. Premise ( Simple conjunction ) : contain
$n$ individual Propositional variables Of Simple conjunction ;
2. The number of occurrence of propositional variables : Each propositional variable all With written words Of form In which , And Only appears once ;
3. Where the propositional variable appears :
The first $\leq i \leq n1≤i≤n ) Words appear in From the left The first i i$
$i$ A place ;
4. Summary of minor items : Meeting the above three conditions Simple conjunction , be called minterm ;
5. $m_imi And M i M_i$
$M_{i}$ The relationship between :①

( 2 ) minterm explain

About minterm Of explain :

1. Number of minimal terms : $2^n$
$2^{n}$ individual minterm ;
2. Not equal to each other : $2^n$
$2^{n}$ A minimal term all Not equal to each other ;
3. minterm : $m_imi Express The first i i$
$i$ A minimal term , among
$i$ Is the minimum term True assignment Of Decimal means ;
4. Minimal item name : The first $m_i$
$m_{i}$ ;

( 3 ) Two propositional variables Of minterm

Two propositional variables

p, q

$p, q$ Of minterm :

1. Write first minterm name : from $m_0, m_1, m_2, m_3$
$m_{0}, m_{1}, m_{2}, m_{3}$ ;
2. Then write the true assignment :
$00, 01, 10, 11$ ;
3. Finally, write the formula ( Simple conjunction ) :
- ① Formula form : The formula is simple conjunctive , $\land q$
  $p \land q$ , among Each propositional variable
  $\neg$ ;
- ② Satisfy the real assignment : The formula needs to meet Its Above
  $00, 01, 10, 11$ Assignment is true assignment , That is, assign values according to reality , Deduce its formula ;
- ③ analysis : True assignment by $\land∧ Both sides should be really , The assignment is 0 , that Corresponding propositional variables Take it with you ¬ \lnot$
  $\neg$ Symbol ;
- ④ Corresponding : Anyone who $0 0 ¬ \lnot Symbol ; Anyone who 1 1$
  $0$ The assignment of , belt
  $1$ The assignment of , Corresponding normal Propositional variables ;

The formula	True assignment	name
$\lnot p \land \lnot q¬p∧¬q$	$\quad 000$	$m_0m0$
$\lnot p \land q¬p∧q$	$\quad 101$	$m_1m1$
$\land \lnot qp∧¬q$	$\quad 010$	$m_2m2$
$\land qp∧q$	$\quad 111$	$m_3m3$

( 4 ) Three propositional variables Of minterm

Three propositional variables

p, q, r

$p, q, r$ Of minterm :

1. Write first minterm name : from $m_0, m_1, m_2, m_3, m_4, m_5, m_6, m_7$
$m_{0}, m_{1}, m_{2}, m_{3}, m_{4}, m_{5}, m_{6}, m_{7}$ ;
2. Then write the true assignment :
$000, 001, 010, 011, 100, 101, 110, 111$ ;
3. Finally, write the formula ( Simple conjunction ) :
- ① Formula form : The formula is simple conjunctive , $\land q \land r$
  $p \land q \land r$ , among Each propositional variable
  $\neg$ ;
- ② Satisfy the real assignment : The formula needs to meet Its Above
  $000, 001, 010, 011, 100, 101, 110, 111$ Assignment is true assignment , That is, assign values according to reality , Deduce its formula ;
- ③ analysis : True assignment by $\lnot$
  $\neg$ Symbol ;
- ④ Corresponding : Anyone who $0 0 ¬ \lnot Symbol ; Anyone who 1 1$
  $0$ The assignment of , belt
  $1$ The assignment of , Corresponding normal Propositional variables ;

The formula	True assignment	name
$\lnot p \land \lnot q \land \lnot r¬p∧¬q∧¬r$	$\quad 0 \quad 0000$	$m_0m0$
$\lnot p \land \lnot q \land r¬p∧¬q∧r$	$\quad 0 \quad 1001$	$m_1m1$
$\lnot p \land q \land \lnot r¬p∧q∧¬r$	$\quad 1 \quad 0010$	$m_2m2$
$\lnot p \land q \land r¬p∧q∧r$	$\quad 1 \quad 1011$	$m_3m3$
$\land \lnot q \land \lnot rp∧¬q∧¬r$	$\quad 0 \quad 0100$	$m_4m4$
$\land \lnot q \land rp∧¬q∧r$	$\quad 0 \quad 1101$	$m_5m5$
$\land q \land \lnot rp∧q∧¬r$	$\quad 1 \quad 0110$	$m_6m6$
$\land q \land rp∧q∧r$	$\quad 1 \quad 1111$	$m_7m7$

( 5 ) minterm True assignment The formula name Between Of conversion And Deduce

minterm True assignment The formula name Between Of conversion And Deduce :

1. True assignment To The formula Deduction between : The formula Of True assignment list , It's true assignment ; Assign values according to reality Write The formula , 0 Corresponding Propositional variables belt no $\lnot$
$\neg$ , 1 Corresponding Normal propositional variables ;
2. name To True assignment Between Deduce : This The most simple , Direct will Subscript It's written in Binary form that will do ;
3. The formula To name Between Deduce : Direct deduction More difficult , Must pass True assignment Make a transition , Write first True assignment , Then think of it as Binary number To Decimal subscript is enough ;

3. The maximal term

( 1 ) The maximal term brief introduction

The maximal term : The maximal term yes A kind of Simple disjunctive ;

1. Premise ( Simple disjunctive ) : contain
$n$ individual Propositional variables Of Simple disjunctive ;
2. The number of occurrence of propositional variables : Each propositional variable all With written words Of form In which , And Only appears once ;
3. Where the propositional variable appears :
The first $\leq i \leq n1≤i≤n ) Words appear in From the left The first i i$
$i$ A place ;
4. Summary of maximal items : Meeting the above three conditions Simple disjunctive , be called The maximal term ;

( 2 ) The maximal term explain

About The maximal term Of explain :

1. The number of maximal terms : $2^n$
$2^{n}$ individual The maximal term ;
2. Not equal to each other : $2^n$
$2^{n}$ A very large item all Not equal to each other ;
3. The maximal term : $m_imi Express The first i i$
$i$ A very large item , among
$i$ Is the largest item False assignment Of Decimal means ;
4. Maximum item name : The first $M_i$
$M_{i}$ ;
5. $m_imi And M i M_i$
$M_{i}$ The relationship between :①

( 3 ) The maximal term of two propositional variables

Two propositional variables

p, q

$p, q$ Of The maximal term :

1. Write first The maximal term name : from $M_0, M_1, M_2, M_3$
$M_{0}, M_{1}, M_{2}, M_{3}$ ;
2. Then write it as a false assignment :
$00, 01, 10, 11$ ;
3. Finally, write the formula ( Simple disjunctive ) :
- ① Formula form : The formula is a simple disjunctive , $\land q$
  $p \land q$ , among Each propositional variable
  $\neg$ ;
- ② Satisfy false assignment : The formula needs to meet Its Above
  $00, 01, 10, 11$ Assignment is false assignment , That is, assign value according to false , Deduce its formula ;
- ③ analysis : False assignment by $\land∧ Both sides should be false , The assignment is 0 , Then the corresponding propositional variable is Normal propositional variables , No negative sign ¬ \lnot$
  $\neg$ ;
- ④ Corresponding : Anyone who $1 1 ¬ \lnot Symbol ; Anyone who 0 0$
  $1$ The assignment of , belt
  $0$ The assignment of , Corresponding normal Propositional variables ;

The formula	False assignment	name
$\lor qp∨q$	$\quad 000$	$M_0M0$
$\lor \lnot qp∨¬q$	$\quad 101$	$M_1M1$
$\lnot p \lor q¬p∨q$	$\quad 010$	$M_2M2$
$\lnot p \lor \lnot q¬p∨¬q$	$\quad 111$	$M_3M3$

( 4 ) The maximal term of three propositional variables

Three propositional variables

p, q, r

$p, q, r$ Of The maximal term :

1. Write first The maximal term name : from $M_0, M_1, M_2, M_3, M_4, M_5, M_6, M_7$
$M_{0}, M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}, M_{7}$ ;
2. Then write it as a false assignment :
$000, 001, 010, 011, 100, 101, 110, 111$ ;
3. Finally, write the formula ( Simple disjunctive ) :
- ① Formula form : The formula is a simple disjunctive , $\land q \land r$
  $p \land q \land r$ , among Each propositional variable
  $\neg$ ;
- ② Satisfy false assignment : The formula needs to meet Its Above
  $000, 001, 010, 011, 100, 101, 110, 111$ Assignment is false assignment , That is, assign values according to reality , Deduce its formula ;
- ③ analysis : False assignment by $\lnot$
  $\neg$ ;
- ④ Corresponding : Anyone who $1 1 ¬ \lnot Symbol ; Anyone who 0 0$
  $1$ The assignment of , belt
  $0$ The assignment of , Corresponding normal Propositional variables ;

The formula	False assignment	name
$\lor q \lor rp∨q∨r$	$\quad 0 \quad 0000$	$M_0M0$
$\lor q \lor \lnot rp∨q∨¬r$	$\quad 0 \quad 1001$	$M_1M1$
$\lor \lnot q \lor rp∨¬q∨r$	$\quad 1 \quad 0010$	$M_2M2$
$\lor \lnot q \lor \lnot rp∨¬q∨¬r$	$\quad 1 \quad 1011$	$M_3M3$
$\lnot p \lor q \lor r¬p∨q∨r$	$\quad 0 \quad 0100$	$M_4M4$
$\lnot p \lor q \lor \lnot r¬p∨q∨¬r$	$\quad 0 \quad 1101$	$M_5M5$
$\lnot p \lor \lnot q \lor r¬p∨¬q∨r$	$\quad 1 \quad 0110$	$M_6M6$
$\lnot p \lor \lnot q \lor \lnot r¬p∨¬q∨¬r$	$\quad 1 \quad 1111$	$M_7M7$

( 5 ) The maximal term False assignment The formula name Between Of conversion And Deduce

The maximal term False assignment The formula name Between Of conversion And Deduce :

1. False assignment To The formula Deduction between : The formula Of False assignment list , Is false assignment ; Assign value according to false Write The formula , $\lnot¬ , 0 0$
$0$ Corresponding Normal propositional variables ;
2. name To False assignment Between Deduce : This The most simple , Direct will Subscript It's written in Binary form that will do ;
3. The formula To name Between Deduce : Direct deduction More difficult , Must pass False assignment Make a transition , Write first False assignment , Then think of it as Binary number To Decimal subscript is enough ;

Two . title

1. Use equivalent calculus to find Principal disjunctive normal form and The main conjunctive paradigm

subject : Use equivalent calculus to find Principal disjunctive normal form and The main conjunctive paradigm ;

Conditions : $\rightarrow \lnot q) \rightarrow r$
$A = (p \to \neg q) \to r$
problem 1 : seek Principal disjunctive normal form and The LORD takes normal form ;

answer :

① step One : Find a conjunctive normal form :

(

→

)

→

(p \rightarrow \lnot q) \rightarrow r

$(p \to \neg q) \to r$

( Use implication equivalence :

→

∨

A \rightarrow B \iff \lnot A \lor B

$A \to B * \neg A \lor B$ , eliminate The outer Implication symbol )

(

→

)

∨

\iff \lnot (p \rightarrow \lnot q) \lor r

$* \neg (p \to \neg q) \lor r$

( Use implication equivalence :

→

∨

A \rightarrow B \iff \lnot A \lor B

$A \to B * \neg A \lor B$ , Eliminate the inner Implication symbol )

(

∨

)

∨

\iff \lnot (\lnot p \lor \lnot q) \lor r

$* \neg (\neg p \lor \neg q) \lor r$

( Use De Morgan's law :

(

∨

)

∧

\lnot (A \lor B) \iff \lnot A \land \lnot B

$\neg (A \lor B) * \neg A \land \neg B$ , Handle

(

∨

)

\lnot (\lnot p \lor \lnot q)

$\neg (\neg p \lor \neg q)$ part )

(

∧

)

∨

\iff ( p \land q) \lor r

$* (p \land q) \lor r$

( Use exchange rate :

∨

A \lor B \iff B \lor A

$A \lor B * B \lor A$ )

∨

(

∧

)

\iff r \lor ( p \land q)

$* r \lor (p \land q)$

( Use allocation rate :

∨

(

∧

)

(

∨

)

∧

(

∨

)

A \lor (B \land C) \iff (A \lor B) \land (A \lor C)

$A \lor (B \land C) * (A \lor B) \land (A \lor C)$ )

(

∨

)

∧

(

∨

)

\iff (r \lor p) \land (r \lor q)

$* (r \lor p) \land (r \lor q)$

( Use exchange rate :

∨

A \lor B \iff B \lor A

$A \lor B * B \lor A$ )

(

∨

)

∧

(

∨

)

\iff (p \lor r) \land (q \lor r)

$* (p \lor r) \land (q \lor r)$

Current situation analysis :

1> Conjunctive paradigm : here , $\lor r) \land (q \lor r)$
$(p \lor r) \land (q \lor r)$ It is a conjunctive paradigm , According to the conjunctive paradigm Ask the Lord to take normal form ;
2> Split : Separately $\lor r)(p∨r) and ( q ∨ r ) (q \lor r)$
$(q \lor r)$ To The maximal term ;

② Step two : take

(

∨

)

(p \lor r)

$(p \lor r)$ To The main conjunctive paradigm :

(

∨

)

(p \lor r)

$(p \lor r)$

( Use Law of zero :

∨

A \lor 0 \iff A

$A \lor 0 * A$ , disjunction , Disjunction one

$0$ after , Its value remains unchanged )

(

∨

)

\iff (p \lor 0 \lor r)

$* (p \lor 0 \lor r)$

( Use Law of contradiction :

∧

A \land A = 0

$A \land A = 0$ , introduce Propositional argument

$q$ , That is to use

∧

A \land A

$A \land A$ Replace In the formula

$0$ )

(

∨

(

∧

)

∨

)

\iff (p \lor ( q \land \lnot q ) \lor r)

$* (p \lor (q \land \neg q) \lor r)$

( Use the commutative law

∨

A \lor B \iff B \lor A

$A \lor B * B \lor A$ and Associative law

(

∨

)

∨

(

∨

)

(A \lor B) \lor C \iff A \lor (B \lor C)

$(A \lor B) \lor C * A \lor (B \lor C)$ )

(

∨

)

∨

(

∧

)

\iff ( ( p \lor r ) \lor ( q \land \lnot q ) )

$* ((p \lor r) \lor (q \land \neg q))$

( Use the law of distribution :

∨

(

∧

)

(

∧

)

∨

(

∧

)

A \lor (B \land C) \iff (A \land B) \lor (A \land C)

$A \lor (B \land C) * (A \land B) \lor (A \land C)$ , take

p,q,r

$p, q, r$ All assembled into a disjunctive )

(

∨

)

∧

(

∨

)

\iff (p \lor r \lor q) \land (p \lor r \lor \lnot q)

$* (p \lor r \lor q) \land (p \lor r \lor \neg q)$

( Use the commutative law )

(

∨

)

∧

(

∨

)

\iff (p \lor q \lor r) \land (p \lor \lnot q \lor r)

$* (p \lor q \lor r) \land (p \lor \neg q \lor r)$

according to The maximal term The formula Write the corresponding serial number :

1> $\lor q \lor r)$
$(p \lor q \lor r)$ : False assignment
2> $\lor \lnot q \lor r)$
$(p \lor \neg q \lor r)$ : False assignment
3> $\lor r)$
$(p \lor r)$ Corresponding The principal conjunctive paradigm is :

③ Step three : take

(

∨

)

(q \lor r)

$(q \lor r)$ To The main conjunctive paradigm :

(

∨

)

(q \lor r)

$(q \lor r)$

( Use Law of zero :

∨

A \lor 0 \iff A

$A \lor 0 * A$ , disjunction , Disjunction one

$0$ after , Its value remains unchanged )

(

∨

)

\iff (0 \lor q \lor r)

$* (0 \lor q \lor r)$

( Use Law of contradiction :

∧

A \land A = 0

$A \land A = 0$ , introduce Propositional argument

$q$ , That is to use

∧

A \land A

$A \land A$ Replace In the formula

$0$ )

(

∧

)

∨

)

\iff (( p \land \lnot p ) \lor q \lor r)

$* ((p \land \neg p) \lor q \lor r)$

( Use the law of distribution :

∨

(

∧

)

(

∧

)

∨

(

∧

)

A \lor (B \land C) \iff (A \land B) \lor (A \land C)

$A \lor (B \land C) * (A \land B) \lor (A \land C)$ , take

p,q,r

$p, q, r$ All assembled into a disjunctive )

(

∨

)

∧

(

∨

)

\iff (p \lor r \lor q) \land (\lnot p \lor r \lor q)

$* (p \lor r \lor q) \land (\neg p \lor r \lor q)$

according to The maximal term The formula Write the corresponding serial number :

1> $\lor q \lor r)$
$(p \lor q \lor r)$ : False assignment
2> $(\lnot p \lor q \lor r)$
$(\neg p \lor q \lor r)$ : False assignment
3> $\lor r)$
$(p \lor r)$ Corresponding The principal conjunctive paradigm is :

The final result of this topic :

(

→

)

(p \rightarrow \lnot q)

$(p \to \neg q)$

( Step one Conclusion )

(

∨

)

∧

(

∨

)

\iff (p \lor r) \land (q \lor r)

$* (p \lor r) \land (q \lor r)$

( Move step two and Step three The result is substituted into the above formula )

(

∧

)

∧

(

∧

)

\iff (M_0 \land M_2) \land (M_0 \land M_4)

$* (M_{0} \land M_{2}) \land (M_{0} \land M_{4})$

( According to the law of Union You can eliminate the brackets take

∧

M_0 \land M_0

$M_{0} \land M_{0}$ combined )

(

∧

)

∧

\iff ( M_0 \land M_0 ) \land M_2 \land M_4

$* (M_{0} \land M_{0}) \land M_{2} \land M_{4}$

( according to Idempotent law :

∧

A \land A \iff A

$A \land A * A$ , Can be eliminated One

M_0

$M_{0}$ )

∧

\iff M_0 \land M_2 \land M_4

$* M_{0} \land M_{2} \land M_{4}$

2. Use Truth table method seek Principal disjunctive normal form and The main conjunctive paradigm

subject : Use Truth table method seek Principal disjunctive normal form and The main conjunctive paradigm ;

Conditions : $\rightarrow \lnot q) \rightarrow r$
$A = (p \to \neg q) \to r$
problem 1 : seek Principal disjunctive normal form and The LORD takes normal form ;

answer :

① First, list the truth table ( The more detailed the truth table of the column, the better , Miscalculation several times )

$\quad q \quad rpqr$	minterm	The maximal term
$\quad 0 \quad 0000$	$m_0m0$	$M_0$ $M_{0}$
$\quad 0 \quad 1001$	$m_1$ $m_{1}$	$M_1M1$
$\quad 1 \quad 0010$	$m_2m2$	$M_2$ $M_{2}$
$\quad 1 \quad 1011$	$m_3$ $m_{3}$	$M_3M3$
$\quad 0 \quad 0100$	$m_4m4$	$M_4$ $M_{4}$
$\quad 0 \quad 1101$	$m_5$ $m_{5}$	$M_5M5$
$\quad 1 \quad 0110$	$m_6$ $m_{6}$	$M_6M6$
$\quad 1 \quad 1111$	$m_7$ $m_{7}$	$M_7M7$