Show that [¬p ∧ (p ∨ q)] → q is a tautology [closed]

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How can I show that [¬p ∧ (p ∨ q)] → q is a tautology by using the logical equivalences?

discrete-mathematics logic proof-writing

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asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

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closed as off-topic by John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus Jan 21 at 4:02

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How can I show that [¬p ∧ (p ∨ q)] → q is a tautology by using the logical equivalences?

discrete-mathematics logic proof-writing

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asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

$endgroup$

closed as off-topic by John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus Jan 21 at 4:02

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." – John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus

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

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$begingroup$

How can I show that [¬p ∧ (p ∨ q)] → q is a tautology by using the logical equivalences?

discrete-mathematics logic proof-writing

share|cite|improve this question

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

$endgroup$

How can I show that [¬p ∧ (p ∨ q)] → q is a tautology by using the logical equivalences?

discrete-mathematics logic proof-writing

discrete-mathematics logic proof-writing

share|cite|improve this question

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

share|cite|improve this question

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

share|cite|improve this question

share|cite|improve this question

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

asked Jan 21 at 1:07

Usama GhawjiUsama Ghawji

465

465

closed as off-topic by John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus Jan 21 at 4:02

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." – John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus

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

closed as off-topic by John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus Jan 21 at 4:02

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." – John Douma, Shailesh, Eric Wofsey, max_zorn, Leucippus

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

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2 Answers
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You can do this via "basic" facts like De Morgan Laws for logic.

$$[neg p wedge(pvee q)]implies q equiv$$
$$neg [neg p wedge(pvee q)] vee qequiv$$
$$pvee neg (pvee q)vee qequiv$$
$$p vee neg p wedge neg q vee q equiv$$
$$(p vee neg p) wedge (neg q vee q)equiv$$
$$T wedge Tequiv T$$

Hence, the statement is a tautology.

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answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

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One approach is to do a truth table. For each combination of p and q, 1 and 0, figure out first (p v q). Then you take that answer and figure out the neg. p and the previous answer. Then lastly, that answer implies q. The final column for the truth table should be all 1 or true. Hopefully, you could understand my poorly written answer.

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answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

You can do this via "basic" facts like De Morgan Laws for logic.

$$[neg p wedge(pvee q)]implies q equiv$$
$$neg [neg p wedge(pvee q)] vee qequiv$$
$$pvee neg (pvee q)vee qequiv$$
$$p vee neg p wedge neg q vee q equiv$$
$$(p vee neg p) wedge (neg q vee q)equiv$$
$$T wedge Tequiv T$$

Hence, the statement is a tautology.

share|cite|improve this answer

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

$endgroup$

add a comment | 

1

$begingroup$

You can do this via "basic" facts like De Morgan Laws for logic.

$$[neg p wedge(pvee q)]implies q equiv$$
$$neg [neg p wedge(pvee q)] vee qequiv$$
$$pvee neg (pvee q)vee qequiv$$
$$p vee neg p wedge neg q vee q equiv$$
$$(p vee neg p) wedge (neg q vee q)equiv$$
$$T wedge Tequiv T$$

Hence, the statement is a tautology.

share|cite|improve this answer

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

$endgroup$

add a comment | 

1

1

1

$begingroup$

You can do this via "basic" facts like De Morgan Laws for logic.

$$[neg p wedge(pvee q)]implies q equiv$$
$$neg [neg p wedge(pvee q)] vee qequiv$$
$$pvee neg (pvee q)vee qequiv$$
$$p vee neg p wedge neg q vee q equiv$$
$$(p vee neg p) wedge (neg q vee q)equiv$$
$$T wedge Tequiv T$$

Hence, the statement is a tautology.

share|cite|improve this answer

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

$endgroup$

You can do this via "basic" facts like De Morgan Laws for logic.

$$[neg p wedge(pvee q)]implies q equiv$$
$$neg [neg p wedge(pvee q)] vee qequiv$$
$$pvee neg (pvee q)vee qequiv$$
$$p vee neg p wedge neg q vee q equiv$$
$$(p vee neg p) wedge (neg q vee q)equiv$$
$$T wedge Tequiv T$$

Hence, the statement is a tautology.

share|cite|improve this answer

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

answered Jan 21 at 2:50

Evan William ChandraEvan William Chandra

628313

628313

add a comment | 

add a comment | 

0

$begingroup$

One approach is to do a truth table. For each combination of p and q, 1 and 0, figure out first (p v q). Then you take that answer and figure out the neg. p and the previous answer. Then lastly, that answer implies q. The final column for the truth table should be all 1 or true. Hopefully, you could understand my poorly written answer.

share|cite|improve this answer

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

$endgroup$

add a comment | 

0

$begingroup$

One approach is to do a truth table. For each combination of p and q, 1 and 0, figure out first (p v q). Then you take that answer and figure out the neg. p and the previous answer. Then lastly, that answer implies q. The final column for the truth table should be all 1 or true. Hopefully, you could understand my poorly written answer.

share|cite|improve this answer

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

$endgroup$

add a comment | 

0

0

0

$begingroup$

One approach is to do a truth table. For each combination of p and q, 1 and 0, figure out first (p v q). Then you take that answer and figure out the neg. p and the previous answer. Then lastly, that answer implies q. The final column for the truth table should be all 1 or true. Hopefully, you could understand my poorly written answer.

share|cite|improve this answer

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

$endgroup$

One approach is to do a truth table. For each combination of p and q, 1 and 0, figure out first (p v q). Then you take that answer and figure out the neg. p and the previous answer. Then lastly, that answer implies q. The final column for the truth table should be all 1 or true. Hopefully, you could understand my poorly written answer.

share|cite|improve this answer

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

share|cite|improve this answer

share|cite|improve this answer

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

answered Jan 21 at 1:21

Jeffrey NgJeffrey Ng

82

82

add a comment | 

add a comment | 

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