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Mathematical Logic, truth tables, logical equivalence example ( Enter your problem )
  1. Examples
Other related methods
  1. Definitions
  2. Laws of logical connectives
  3. Prepare the truth table
  4. logical validity of the argument

3. Prepare the truth table
(Previous method)

1. Examples





The implication that statement `S` follows from statements `S_1,S_2,S_3,...,S_n` is called an argument.
Thus the statements `S_1^^S_2^^S_3^^...^^S_n=>S` is an argument.
In symbols argument is represented by `(S_1,S_2,S_3,...,S_n;S)`. In this argument `S_1,S_2,S_3,...,S_n` is called hypothesis and S is called conclusion.
Such an argument is said to be logically valid if the conclusion `S` is true whenever the hypothesis statements `S_1,S_2,S_3,...,S_n` are all true. Not all arguments are logical.
To test the logical validity of argument `(S_1,S_2,S_3,...,S_n;S)`, we must form a truth table.
The rows in which all statements `S_1,S_2,S_3,...,S_n` have truth value T int the truth table are called critical rows.
If the conclusion `S` has truth value T in every critical rows of the truth table of an argument then we say that the argument is logically valid.


1. Examine the logical validity of the following argument
Hypothesis : p or q; not p and Conclusion : q


Solution:
Hypothesis :
`S_1:pvvq`

`S_2:~p`

Conclusion :
`S : q`


`(1)``(2)``(3)=(1)vv(2)``(4)=~(1)``(5)`
`p``q``S_1`
`pvvq`
`S_2`
`~p`
`S`
`q`
TTTFT
TFTFF
FTTTcritical rowT
FFFTF


The conclusion `(S)` is true in all critical rows. So the argument is logically valid.


2. Examine the logical validity of the following argument
Hypothesis : (p and not(q)) => r;p or q;q => p and Conclusion : r


Solution:
Hypothesis :
`S_1:(p^^~q)=>r`

`S_2:pvvq`

`S_3:q=>p`

Conclusion :
`S : r`


`(1)``(2)``(3)``(4)=~(2)``(5)=(1)^^(4)``(6)=(5)=>(3)``(7)=(1)vv(2)``(8)=(2)=>(1)``(9)`
`p``q``r``~q``p^^~q``S_1`
`(p^^~q)=>r`
`S_2`
`pvvq`
`S_3`
`q=>p`
`S`
`r`
TTTFFTTTcritical rowT
TTFFFTTTcritical rowF
TFTTTTTTcritical rowT
TFFTTFTTF
FTTFFTTFT
FTFFFTTFF
FFTTFTFTT
FFFTFTFTF


The conclusion `(S)` is not true in all critical rows. So the argument is not logically valid.


3. Examine the logical validity of the following argument
Hypothesis : p => q;q => r and Conclusion : p => r


Solution:
Hypothesis :
`S_1:p=>q`

`S_2:q=>r`

Conclusion :
`S : p=>r`


`(1)``(2)``(3)``(4)=(1)=>(2)``(5)=(2)=>(3)``(6)=(1)=>(3)`
`p``q``r``S_1`
`p=>q`
`S_2`
`q=>r`
`S`
`p=>r`
TTTTTcritical rowT
TTFTFF
TFTFTT
TFFFTF
FTTTTcritical rowT
FTFTFT
FFTTTcritical rowT
FFFTTcritical rowT


The conclusion `(S)` is true in all critical rows. So the argument is logically valid.


4. Examine the logical validity of the following argument
Hypothesis : p => q;p => r and Conclusion : p => (q and r)


Solution:
Hypothesis :
`S_1:p=>q`

`S_2:p=>r`

Conclusion :
`S : p=>(q^^r)`


`(1)``(2)``(3)``(4)=(1)=>(2)``(5)=(1)=>(3)``(6)=(2)^^(3)``(7)=(1)=>(6)`
`p``q``r``S_1`
`p=>q`
`S_2`
`p=>r`
`q^^r``S`
`p=>(q^^r)`
TTTTTTcritical rowT
TTFTFFF
TFTFTFF
TFFFFFF
FTTTTTcritical rowT
FTFTTFcritical rowT
FFTTTFcritical rowT
FFFTTFcritical rowT


The conclusion `(S)` is true in all critical rows. So the argument is logically valid.





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3. Prepare the truth table
(Previous method)





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