How is $[P text AND (Q text OR R)] text IFF [(P text AND Q) text OR (P text AND R)]$ valid?

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

The image below was taken from some reading materials in a public course on the mathematics of computer science. The text states that the proposition [P AND (Q OR R)] IFF [(P AND Q) OR (P AND R)] is valid.

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

logic propositional-calculus

share|cite|improve this question

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

asked Aug 5 at 16:15

sean

1083

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Aug 5 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
  – Henning Makholm
  Aug 5 at 16:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
  – Doug Spoonwood
  Aug 6 at 13:11
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

The image below was taken from some reading materials in a public course on the mathematics of computer science. The text states that the proposition [P AND (Q OR R)] IFF [(P AND Q) OR (P AND R)] is valid.

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

logic propositional-calculus

share|cite|improve this question

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

asked Aug 5 at 16:15

sean

1083

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Aug 5 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
  – Henning Makholm
  Aug 5 at 16:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
  – Doug Spoonwood
  Aug 6 at 13:11
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The image below was taken from some reading materials in a public course on the mathematics of computer science. The text states that the proposition [P AND (Q OR R)] IFF [(P AND Q) OR (P AND R)] is valid.

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

logic propositional-calculus

share|cite|improve this question

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

asked Aug 5 at 16:15

sean

1083

The image below was taken from some reading materials in a public course on the mathematics of computer science. The text states that the proposition [P AND (Q OR R)] IFF [(P AND Q) OR (P AND R)] is valid.

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

logic propositional-calculus

share|cite|improve this question

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

asked Aug 5 at 16:15

sean

1083

share|cite|improve this question

share|cite|improve this question

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

edited Aug 5 at 23:40

Asaf Karagila♦

292k31403733

292k31403733

asked Aug 5 at 16:15

sean

1083

asked Aug 5 at 16:15

sean

1083

asked Aug 5 at 16:15

sean

1083

1083

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Aug 5 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
  – Henning Makholm
  Aug 5 at 16:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
  – Doug Spoonwood
  Aug 6 at 13:11
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Aug 5 at 16:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
  – Henning Makholm
  Aug 5 at 16:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
  – Doug Spoonwood
  Aug 6 at 13:11
  

  
  
  
  

2

2

Here's a MathJax tutorial :)
– Shaun
Aug 5 at 16:24

Here's a MathJax tutorial :)
– Shaun
Aug 5 at 16:24

@Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
– Henning Makholm
Aug 5 at 16:46

@Shaun: That won't help him much here, given that the notation in his source uses words rather than symbols for the connectives anyway.
– Henning Makholm
Aug 5 at 16:46

This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
– Doug Spoonwood
Aug 6 at 13:11

This question indicates one reason as to why you want to identify the principal connective of a formula. Namely, it helps you figure out whether or not you have computed the right formula.
– Doug Spoonwood
Aug 6 at 13:11

add a comment | 

4 Answers
4

active

oldest

votes

up vote
10
down vote



accepted

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
  – amWhy
  Aug 5 at 16:43
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P land (Q lor R)$$

and

$$(P land Q) lor (P land R)$$

are equivalent.

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

add a comment | 

up vote
0
down vote

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

add a comment | 

up vote
0
down vote

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$deftooleftrightarrow$Not just some cases, the clauses evaluate to false in the same cases.   You see that $Pland(Qlor R)$ is false is exactly when $(Pland Q)lor (Pland R)$ is false.   Likewise for all true cases.

In all cases $Pland (Qlor R)$ and $(Pland Q)lor(Pland R)$ have the same truth value.   They are equivalent.   Then $(Pland (Qlor R))too((Pland Q)lor(Pland R))$ is alwas true because that is what the biconditional means.

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873121%2fhow-is-p-text-and-q-text-or-r-text-iff-p-text-and-q-tex%23new-answer', 'question_page');

);

Post as a guest

Name Email

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
10
down vote



accepted

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
  – amWhy
  Aug 5 at 16:43
  
  

  
  
  
  

add a comment | 

up vote
10
down vote



accepted

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
  – amWhy
  Aug 5 at 16:43
  
  

  
  
  
  

add a comment | 

up vote
10
down vote



accepted

up vote
10
down vote



accepted

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

share|cite|improve this answer

share|cite|improve this answer

answered Aug 5 at 16:19

Henning Makholm

226k16292521

answered Aug 5 at 16:19

Henning Makholm

226k16292521

answered Aug 5 at 16:19

Henning Makholm

226k16292521

226k16292521

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
  – amWhy
  Aug 5 at 16:43
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
  – amWhy
  Aug 5 at 16:43
  
  

  
  
  
  

1

1

Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
– amWhy
Aug 5 at 16:43

Indeed, if the final column would have been included with header $$big(Pland (Qlor R)big) leftrightarrow big((P land Q) lor (P land R)big)$$ the column would contain, below that "header" T, T, T, T, T, T, T, T, because whenever the truth values match on the left and right of $leftrightarrow$ match under any truth value assignment to the variables, then the biconditional is true. And so indeed, the proposition you are asking about, sean, is indeed a tautology. You just missed the most crucial column.
– amWhy
Aug 5 at 16:43

add a comment | 

up vote
2
down vote

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P land (Q lor R)$$

and

$$(P land Q) lor (P land R)$$

are equivalent.

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

add a comment | 

up vote
2
down vote

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P land (Q lor R)$$

and

$$(P land Q) lor (P land R)$$

are equivalent.

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

add a comment | 

up vote
2
down vote

up vote
2
down vote

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P land (Q lor R)$$

and

$$(P land Q) lor (P land R)$$

are equivalent.

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P land (Q lor R)$$

and

$$(P land Q) lor (P land R)$$

are equivalent.

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

share|cite|improve this answer

share|cite|improve this answer

edited Aug 5 at 16:21

amWhy

189k25219431

edited Aug 5 at 16:21

amWhy

189k25219431

edited Aug 5 at 16:21

amWhy

189k25219431

189k25219431

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

answered Aug 5 at 16:19

Siong Thye Goh

78.1k134997

78.1k134997

add a comment | 

add a comment | 

up vote
0
down vote

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

add a comment | 

up vote
0
down vote

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

add a comment | 

up vote
0
down vote

up vote
0
down vote

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

share|cite|improve this answer

share|cite|improve this answer

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

answered Aug 5 at 16:19

Cameron Buie

83.5k771153

83.5k771153

add a comment | 

add a comment | 

up vote
0
down vote

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$deftooleftrightarrow$Not just some cases, the clauses evaluate to false in the same cases.   You see that $Pland(Qlor R)$ is false is exactly when $(Pland Q)lor (Pland R)$ is false.   Likewise for all true cases.

In all cases $Pland (Qlor R)$ and $(Pland Q)lor(Pland R)$ have the same truth value.   They are equivalent.   Then $(Pland (Qlor R))too((Pland Q)lor(Pland R))$ is alwas true because that is what the biconditional means.

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$deftooleftrightarrow$Not just some cases, the clauses evaluate to false in the same cases.   You see that $Pland(Qlor R)$ is false is exactly when $(Pland Q)lor (Pland R)$ is false.   Likewise for all true cases.

In all cases $Pland (Qlor R)$ and $(Pland Q)lor(Pland R)$ have the same truth value.   They are equivalent.   Then $(Pland (Qlor R))too((Pland Q)lor(Pland R))$ is alwas true because that is what the biconditional means.

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

add a comment | 

up vote
0
down vote

up vote
0
down vote

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$deftooleftrightarrow$Not just some cases, the clauses evaluate to false in the same cases.   You see that $Pland(Qlor R)$ is false is exactly when $(Pland Q)lor (Pland R)$ is false.   Likewise for all true cases.

In all cases $Pland (Qlor R)$ and $(Pland Q)lor(Pland R)$ have the same truth value.   They are equivalent.   Then $(Pland (Qlor R))too((Pland Q)lor(Pland R))$ is alwas true because that is what the biconditional means.

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$deftooleftrightarrow$Not just some cases, the clauses evaluate to false in the same cases.   You see that $Pland(Qlor R)$ is false is exactly when $(Pland Q)lor (Pland R)$ is false.   Likewise for all true cases.

In all cases $Pland (Qlor R)$ and $(Pland Q)lor(Pland R)$ have the same truth value.   They are equivalent.   Then $(Pland (Qlor R))too((Pland Q)lor(Pland R))$ is alwas true because that is what the biconditional means.

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

answered Aug 6 at 0:07

Graham Kemp

80.1k43275

80.1k43275

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873121%2fhow-is-p-text-and-q-text-or-r-text-iff-p-text-and-q-tex%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email