When should I use RAA in natural deduction proofs?












2












$begingroup$


I can't understand exactly when should I use RAA (reductio ad absurdum) rule in natural deduction proofs? What situation should "trigger" me to think "Now it's time to use RAA"?






logic proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    When you have no other strategies available... then try RAA.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 2 at 12:09










  • $begingroup$
    Do you include also "non invertible rules" in available strategies?
    $endgroup$
    – Maicake
    Jan 2 at 12:14


















2












$begingroup$


I can't understand exactly when should I use RAA (reductio ad absurdum) rule in natural deduction proofs? What situation should "trigger" me to think "Now it's time to use RAA"?






logic proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    When you have no other strategies available... then try RAA.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 2 at 12:09










  • $begingroup$
    Do you include also "non invertible rules" in available strategies?
    $endgroup$
    – Maicake
    Jan 2 at 12:14
















2












2








2





$begingroup$


I can't understand exactly when should I use RAA (reductio ad absurdum) rule in natural deduction proofs? What situation should "trigger" me to think "Now it's time to use RAA"?






logic proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$




I can't understand exactly when should I use RAA (reductio ad absurdum) rule in natural deduction proofs? What situation should "trigger" me to think "Now it's time to use RAA"?






logic proof-theory natural-deduction formal-proofs




logic proof-theory natural-deduction formal-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 18:48









Taroccoesbrocco

5,64271840




5,64271840










asked Jan 2 at 11:51









MaicakeMaicake

715




715








  • 2




    $begingroup$
    When you have no other strategies available... then try RAA.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 2 at 12:09










  • $begingroup$
    Do you include also "non invertible rules" in available strategies?
    $endgroup$
    – Maicake
    Jan 2 at 12:14
















  • 2




    $begingroup$
    When you have no other strategies available... then try RAA.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 2 at 12:09










  • $begingroup$
    Do you include also "non invertible rules" in available strategies?
    $endgroup$
    – Maicake
    Jan 2 at 12:14










2




2




$begingroup$
When you have no other strategies available... then try RAA.
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 12:09




$begingroup$
When you have no other strategies available... then try RAA.
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 12:09












$begingroup$
Do you include also "non invertible rules" in available strategies?
$endgroup$
– Maicake
Jan 2 at 12:14






$begingroup$
Do you include also "non invertible rules" in available strategies?
$endgroup$
– Maicake
Jan 2 at 12:14












1 Answer
1






active

oldest

votes


















2












$begingroup$


What situation should "trigger" me to think "Now it's time to use RAA"?




I can think of $4$ explicit situations:




  1. Your goal is the negation of something. If you goal is $neg varphi$, assume $varphi$ and see if you can get a contradiction


  2. Your goal is some atomic statement $P$: Assume $neg P$, get a contradiction, get $neg neg P$ using RAA, and finally derive $P$ from $neg neg P$ (classical logics typically have a $neg Elim$ rule for this)


  3. Your goal is a disjunction $varphi lor psi$ ... and you don't have any disjunction that you can work with (if you do have a disjunction, set up a $lor Elim$ on that one.). The nice thing about doing an RAA here is that once you assume $neg ( varphi lor psi)$, you should be able to derive both $neg varphi$ and $neg psi$ (using two RAA proofs themselves!), and now you have some useful stuff to work with on your way to a contradiction.


  4. Your goal is an existential $exists x varphi(x)$ ... and you don't have some other existential to work with (if you do have another existential, set up a $lor Elim$ on that one.). So here the assumption will be $neg exists x varphi(x)$ which, with a bit of work (and probably another RAA inside) should allow you to prove $forall x neg varphi(x)$ ... and that will be useful to try and get to a contradiction given whatever else you have.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
    $endgroup$
    – Maicake
    Jan 3 at 9:47










  • $begingroup$
    Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
    $endgroup$
    – Maicake
    Jan 3 at 12:02










  • $begingroup$
    @Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
    $endgroup$
    – Bram28
    Jan 3 at 14:28












  • $begingroup$
    I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
    $endgroup$
    – Maicake
    Jan 3 at 14:33










  • $begingroup$
    @Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
    $endgroup$
    – Bram28
    Jan 3 at 16:54











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2












$begingroup$


What situation should "trigger" me to think "Now it's time to use RAA"?




I can think of $4$ explicit situations:




  1. Your goal is the negation of something. If you goal is $neg varphi$, assume $varphi$ and see if you can get a contradiction


  2. Your goal is some atomic statement $P$: Assume $neg P$, get a contradiction, get $neg neg P$ using RAA, and finally derive $P$ from $neg neg P$ (classical logics typically have a $neg Elim$ rule for this)


  3. Your goal is a disjunction $varphi lor psi$ ... and you don't have any disjunction that you can work with (if you do have a disjunction, set up a $lor Elim$ on that one.). The nice thing about doing an RAA here is that once you assume $neg ( varphi lor psi)$, you should be able to derive both $neg varphi$ and $neg psi$ (using two RAA proofs themselves!), and now you have some useful stuff to work with on your way to a contradiction.


  4. Your goal is an existential $exists x varphi(x)$ ... and you don't have some other existential to work with (if you do have another existential, set up a $lor Elim$ on that one.). So here the assumption will be $neg exists x varphi(x)$ which, with a bit of work (and probably another RAA inside) should allow you to prove $forall x neg varphi(x)$ ... and that will be useful to try and get to a contradiction given whatever else you have.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
    $endgroup$
    – Maicake
    Jan 3 at 9:47










  • $begingroup$
    Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
    $endgroup$
    – Maicake
    Jan 3 at 12:02










  • $begingroup$
    @Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
    $endgroup$
    – Bram28
    Jan 3 at 14:28












  • $begingroup$
    I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
    $endgroup$
    – Maicake
    Jan 3 at 14:33










  • $begingroup$
    @Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
    $endgroup$
    – Bram28
    Jan 3 at 16:54
















2












$begingroup$


What situation should "trigger" me to think "Now it's time to use RAA"?




I can think of $4$ explicit situations:




  1. Your goal is the negation of something. If you goal is $neg varphi$, assume $varphi$ and see if you can get a contradiction


  2. Your goal is some atomic statement $P$: Assume $neg P$, get a contradiction, get $neg neg P$ using RAA, and finally derive $P$ from $neg neg P$ (classical logics typically have a $neg Elim$ rule for this)


  3. Your goal is a disjunction $varphi lor psi$ ... and you don't have any disjunction that you can work with (if you do have a disjunction, set up a $lor Elim$ on that one.). The nice thing about doing an RAA here is that once you assume $neg ( varphi lor psi)$, you should be able to derive both $neg varphi$ and $neg psi$ (using two RAA proofs themselves!), and now you have some useful stuff to work with on your way to a contradiction.


  4. Your goal is an existential $exists x varphi(x)$ ... and you don't have some other existential to work with (if you do have another existential, set up a $lor Elim$ on that one.). So here the assumption will be $neg exists x varphi(x)$ which, with a bit of work (and probably another RAA inside) should allow you to prove $forall x neg varphi(x)$ ... and that will be useful to try and get to a contradiction given whatever else you have.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
    $endgroup$
    – Maicake
    Jan 3 at 9:47










  • $begingroup$
    Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
    $endgroup$
    – Maicake
    Jan 3 at 12:02










  • $begingroup$
    @Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
    $endgroup$
    – Bram28
    Jan 3 at 14:28












  • $begingroup$
    I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
    $endgroup$
    – Maicake
    Jan 3 at 14:33










  • $begingroup$
    @Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
    $endgroup$
    – Bram28
    Jan 3 at 16:54














2












2








2





$begingroup$


What situation should "trigger" me to think "Now it's time to use RAA"?




I can think of $4$ explicit situations:




  1. Your goal is the negation of something. If you goal is $neg varphi$, assume $varphi$ and see if you can get a contradiction


  2. Your goal is some atomic statement $P$: Assume $neg P$, get a contradiction, get $neg neg P$ using RAA, and finally derive $P$ from $neg neg P$ (classical logics typically have a $neg Elim$ rule for this)


  3. Your goal is a disjunction $varphi lor psi$ ... and you don't have any disjunction that you can work with (if you do have a disjunction, set up a $lor Elim$ on that one.). The nice thing about doing an RAA here is that once you assume $neg ( varphi lor psi)$, you should be able to derive both $neg varphi$ and $neg psi$ (using two RAA proofs themselves!), and now you have some useful stuff to work with on your way to a contradiction.


  4. Your goal is an existential $exists x varphi(x)$ ... and you don't have some other existential to work with (if you do have another existential, set up a $lor Elim$ on that one.). So here the assumption will be $neg exists x varphi(x)$ which, with a bit of work (and probably another RAA inside) should allow you to prove $forall x neg varphi(x)$ ... and that will be useful to try and get to a contradiction given whatever else you have.







share|cite|improve this answer









$endgroup$




What situation should "trigger" me to think "Now it's time to use RAA"?




I can think of $4$ explicit situations:




  1. Your goal is the negation of something. If you goal is $neg varphi$, assume $varphi$ and see if you can get a contradiction


  2. Your goal is some atomic statement $P$: Assume $neg P$, get a contradiction, get $neg neg P$ using RAA, and finally derive $P$ from $neg neg P$ (classical logics typically have a $neg Elim$ rule for this)


  3. Your goal is a disjunction $varphi lor psi$ ... and you don't have any disjunction that you can work with (if you do have a disjunction, set up a $lor Elim$ on that one.). The nice thing about doing an RAA here is that once you assume $neg ( varphi lor psi)$, you should be able to derive both $neg varphi$ and $neg psi$ (using two RAA proofs themselves!), and now you have some useful stuff to work with on your way to a contradiction.


  4. Your goal is an existential $exists x varphi(x)$ ... and you don't have some other existential to work with (if you do have another existential, set up a $lor Elim$ on that one.). So here the assumption will be $neg exists x varphi(x)$ which, with a bit of work (and probably another RAA inside) should allow you to prove $forall x neg varphi(x)$ ... and that will be useful to try and get to a contradiction given whatever else you have.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 20:11









Bram28Bram28

63.2k44793




63.2k44793












  • $begingroup$
    Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
    $endgroup$
    – Maicake
    Jan 3 at 9:47










  • $begingroup$
    Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
    $endgroup$
    – Maicake
    Jan 3 at 12:02










  • $begingroup$
    @Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
    $endgroup$
    – Bram28
    Jan 3 at 14:28












  • $begingroup$
    I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
    $endgroup$
    – Maicake
    Jan 3 at 14:33










  • $begingroup$
    @Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
    $endgroup$
    – Bram28
    Jan 3 at 16:54


















  • $begingroup$
    Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
    $endgroup$
    – Maicake
    Jan 3 at 9:47










  • $begingroup$
    Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
    $endgroup$
    – Maicake
    Jan 3 at 12:02










  • $begingroup$
    @Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
    $endgroup$
    – Bram28
    Jan 3 at 14:28












  • $begingroup$
    I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
    $endgroup$
    – Maicake
    Jan 3 at 14:33










  • $begingroup$
    @Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
    $endgroup$
    – Bram28
    Jan 3 at 16:54
















$begingroup$
Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
$endgroup$
– Maicake
Jan 3 at 9:47




$begingroup$
Great answer thanks. Point 3 remembered me that the exclude middle proof use RAA in this way.
$endgroup$
– Maicake
Jan 3 at 9:47












$begingroup$
Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
$endgroup$
– Maicake
Jan 3 at 12:02




$begingroup$
Can I ask you if this is the case . I have to proof {A => (-B or C), C=> B, D=> B} -> B or -(A and B) but I'm struggling to do it.
$endgroup$
– Maicake
Jan 3 at 12:02












$begingroup$
@Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
$endgroup$
– Bram28
Jan 3 at 14:28






$begingroup$
@Maicake Your goal is a disjunction ... and you don't have a disjunction to work with? Sounds like a job for RAA! That is, assume the negation of that, and see if that leads to a contradiction. Also, from the negation of that you can quickly infer $neg B$ as well as $A land B$ .... from which the contradiction almost immediately follows .... so you're not even using any of the premises ... are you sure you wrote that out correctly?
$endgroup$
– Bram28
Jan 3 at 14:28














$begingroup$
I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
$endgroup$
– Maicake
Jan 3 at 14:33




$begingroup$
I solved it using Exclude Middle rule as first move. How can you infer (using natural deduction steps) infer -B and A / B from -(B / -(A and B)) ?
$endgroup$
– Maicake
Jan 3 at 14:33












$begingroup$
@Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
$endgroup$
– Bram28
Jan 3 at 16:54




$begingroup$
@Maicake For ~B: OK, that's a negation .... sounds like a job for ... RAA! So, assume B. Then by v Intro you get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~B. We can get A&B the exact same way: Assume ~(A &B). By v Intro, you can get B v ~(A & B), and that contradicts ~(B v ~(A & B)). So, by RAA we can conclude ~~(A &B), and by & Elim we get A&B. Yes, RAA is your friend! :)
$endgroup$
– Bram28
Jan 3 at 16:54


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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Gyllenstierna

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