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Given B, ¬(( ¬A ∧ B) ∨ C) Prove A ∧ B (not able to use ana con and can only use taut con for De Morgans equivalency) [closed]

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-7

$begingroup$

enter image description here

Cant use ana con at all and can only use taut con for demorgans equivalency

logic

share|cite|improve this question

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

$endgroup$

closed as off-topic by Leucippus, Paul Frost, Saad, Shailesh, Jyrki Lahtonen Mar 30 at 5:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Paul Frost, Saad, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try to show us what you've tried already; show us effort in your post. It's easier to help out if it's clear what you do and do not understand and in what way you would like the problem solved.
  $endgroup$
  – Marc
  Mar 29 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: You do not need taut con for this.
  $endgroup$
  – Graham Kemp
  Mar 30 at 5:03
  

  
  
  
  

add a comment | 

-7

$begingroup$

enter image description here

Cant use ana con at all and can only use taut con for demorgans equivalency

logic

share|cite|improve this question

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

$endgroup$

closed as off-topic by Leucippus, Paul Frost, Saad, Shailesh, Jyrki Lahtonen Mar 30 at 5:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Leucippus, Paul Frost, Saad, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Try to show us what you've tried already; show us effort in your post. It's easier to help out if it's clear what you do and do not understand and in what way you would like the problem solved.
  $endgroup$
  – Marc
  Mar 29 at 22:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: You do not need taut con for this.
  $endgroup$
  – Graham Kemp
  Mar 30 at 5:03
  

  
  
  
  

add a comment | 

$begingroup$

enter image description here

Cant use ana con at all and can only use taut con for demorgans equivalency

logic

share|cite|improve this question

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

$endgroup$

share|cite|improve this question

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

share|cite|improve this question

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

edited Mar 29 at 23:40

Matt Samuel

39.2k63770

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

asked Mar 29 at 22:36

Libby HejtmancikLibby Hejtmancik

11

1 Answer
1

active

oldest

votes

1

$begingroup$

$B$ is true. $(neg A land B)lor C $ is not true, so both $(neg Aland B)$ and $C$ are not true. As $(neg Aland B)$ is not true, we have either:

• 
  $neg A$ is not true and $B$ is true


• 
  $B$ is not true and $neg A$ is true


• Both $neg A$ and $B$ are not true

The latter two cannot be true as it's given that $B$ is true. The only case that can occur is the first one; $neg A$ is not true so $A$ is true. Given is that $B$ is true, so conclude that $A land B$ is true.

Or in more propositional-logic-like language:

$$B landneg(neg A land B)lor C ) implies B landneg(neg A land B)land neg Cimplies B landneg(neg A land B) implies $$

$$B land (negneg A lor neg B) implies B land negneg A implies Bland A$$

share|cite|improve this answer

edited Mar 29 at 23:01

answered Mar 29 at 22:45

MarcMarc

510211

$endgroup$

add a comment | 

1

$begingroup$

$B$ is true. $(neg A land B)lor C $ is not true, so both $(neg Aland B)$ and $C$ are not true. As $(neg Aland B)$ is not true, we have either:

• 
  $neg A$ is not true and $B$ is true


• 
  $B$ is not true and $neg A$ is true


• Both $neg A$ and $B$ are not true

The latter two cannot be true as it's given that $B$ is true. The only case that can occur is the first one; $neg A$ is not true so $A$ is true. Given is that $B$ is true, so conclude that $A land B$ is true.

Or in more propositional-logic-like language:

$$B landneg(neg A land B)lor C ) implies B landneg(neg A land B)land neg Cimplies B landneg(neg A land B) implies $$

$$B land (negneg A lor neg B) implies B land negneg A implies Bland A$$

share|cite|improve this answer

edited Mar 29 at 23:01

answered Mar 29 at 22:45

MarcMarc

510211

$endgroup$

add a comment | 

1

$begingroup$

$B$ is true. $(neg A land B)lor C $ is not true, so both $(neg Aland B)$ and $C$ are not true. As $(neg Aland B)$ is not true, we have either:

• 
  $neg A$ is not true and $B$ is true


• 
  $B$ is not true and $neg A$ is true


• Both $neg A$ and $B$ are not true

The latter two cannot be true as it's given that $B$ is true. The only case that can occur is the first one; $neg A$ is not true so $A$ is true. Given is that $B$ is true, so conclude that $A land B$ is true.

Or in more propositional-logic-like language:

$$B landneg(neg A land B)lor C ) implies B landneg(neg A land B)land neg Cimplies B landneg(neg A land B) implies $$

$$B land (negneg A lor neg B) implies B land negneg A implies Bland A$$

share|cite|improve this answer

edited Mar 29 at 23:01

answered Mar 29 at 22:45

MarcMarc

510211

$endgroup$

add a comment | 

$begingroup$

$B$ is true. $(neg A land B)lor C $ is not true, so both $(neg Aland B)$ and $C$ are not true. As $(neg Aland B)$ is not true, we have either:

• 
  $neg A$ is not true and $B$ is true


• 
  $B$ is not true and $neg A$ is true


• Both $neg A$ and $B$ are not true

The latter two cannot be true as it's given that $B$ is true. The only case that can occur is the first one; $neg A$ is not true so $A$ is true. Given is that $B$ is true, so conclude that $A land B$ is true.

Or in more propositional-logic-like language:

$$B landneg(neg A land B)lor C ) implies B landneg(neg A land B)land neg Cimplies B landneg(neg A land B) implies $$

$$B land (negneg A lor neg B) implies B land negneg A implies Bland A$$

share|cite|improve this answer

edited Mar 29 at 23:01

answered Mar 29 at 22:45

MarcMarc

510211

$endgroup$

share|cite|improve this answer

edited Mar 29 at 23:01

answered Mar 29 at 22:45

MarcMarc

510211

answered Mar 29 at 22:45

MarcMarc

510211

answered Mar 29 at 22:45

MarcMarc

510211