Mixing
Natural deduction (Logic) proof help
$begingroup$
I'm very new to natural deduction and have been stuck trying to prove this argument all day:
$Ato ¬B,$
$¬Bto ¬C,$
Therefore, $Cto ¬A$.
I've been told I need to use modus tollens on the first premise, but I am not sure what to use it on/what to assume, as I thought using modus tollens on premise 1 would yield $B$, which I don't need, instead of $lnot A.$
Thank you.
logic propositional-calculus
edited Jan 17 at 12:56
jordan_glen
1
asked Jan 17 at 11:13
ashalrikashalrik
62
$endgroup$
add a comment |
$begingroup$
I'm very new to natural deduction and have been stuck trying to prove this argument all day:
$Ato ¬B,$
$¬Bto ¬C,$
Therefore, $Cto ¬A$.
I've been told I need to use modus tollens on the first premise, but I am not sure what to use it on/what to assume, as I thought using modus tollens on premise 1 would yield $B$, which I don't need, instead of $lnot A.$
Thank you.
logic propositional-calculus
edited Jan 17 at 12:56
jordan_glen
1
asked Jan 17 at 11:13
ashalrikashalrik
62
$endgroup$
1
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29
add a comment |
$begingroup$
I'm very new to natural deduction and have been stuck trying to prove this argument all day:
$Ato ¬B,$
$¬Bto ¬C,$
Therefore, $Cto ¬A$.
I've been told I need to use modus tollens on the first premise, but I am not sure what to use it on/what to assume, as I thought using modus tollens on premise 1 would yield $B$, which I don't need, instead of $lnot A.$
Thank you.
logic propositional-calculus
edited Jan 17 at 12:56
jordan_glen
1
asked Jan 17 at 11:13
ashalrikashalrik
62
$endgroup$
I'm very new to natural deduction and have been stuck trying to prove this argument all day:
$Ato ¬B,$
$¬Bto ¬C,$
Therefore, $Cto ¬A$.
I've been told I need to use modus tollens on the first premise, but I am not sure what to use it on/what to assume, as I thought using modus tollens on premise 1 would yield $B$, which I don't need, instead of $lnot A.$
Thank you.
logic propositional-calculus
logic propositional-calculus
edited Jan 17 at 12:56
jordan_glen
1
asked Jan 17 at 11:13
ashalrikashalrik
62
edited Jan 17 at 12:56
jordan_glen
1
asked Jan 17 at 11:13
ashalrikashalrik
62
edited Jan 17 at 12:56
jordan_glen
1
edited Jan 17 at 12:56
jordan_glen
1
edited Jan 17 at 12:56
jordan_glen
1
1
asked Jan 17 at 11:13
ashalrikashalrik
62
asked Jan 17 at 11:13
ashalrikashalrik
62
asked Jan 17 at 11:13
ashalrikashalrik
62
62
1
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29
add a comment |
1
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29
1
1
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming the availability of Modus Tollens rule, we have :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $lnot C$ --- assumed [b]
5) $lnot lnot C$ --- from 3) and 4) by $lnot$elim followed by $lnot$-intro (also called Double Negation introduction), discharging [b]
6) $lnot lnot B$ --- from 5) and 2) by MT
7) $lnot A$ --- from 6) and 1) by MT
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
Note : without MT the proof is quite simple :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $lnot B$ --- from 4) and 1) by $to$-elim
6) $lnot C$ --- from 5) and 2) by $to$-elim
7) $lnot A$ --- from 4) with 3) and 6) by $lnot$-elim followed by $lnot$-intro, discharging [b]
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
add a comment |
$begingroup$
If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.
Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$displaystyle Pto Q$", and "$displaystyle Q to R$", appear on lines of a proof, "$displaystyle Pto R$" can be placed on a subsequent line.
In your argument, we then have
$(1) ;;Ato lnot B,qquad$ (premise)
$(2) ;;lnot B to lnot C,,quad$ (premise)
$(3) ;;Ato lnot C,qquad$ HS, $(1), (2)$
$(4) ;;C to lnot A,qquad$ contraposive of $(3).$
Note that instead of relying on the equivalence of an implication $pto q$ with its contrapositive, $lnot q to lnot p$, we can proceed from $(3)$ to
$(4) ;;quad C quad$ assumption
$(5) ;;quad lnotlnot Cquad$ double negation on $(4)$
$(6) ;;quad lnot Aquad$ ($3$, $5$, modus tollens)
$(7) ;; Cto lnot A quad$ ($4-6$)
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
$endgroup$
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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076856%2fnatural-deduction-logic-proof-help%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming the availability of Modus Tollens rule, we have :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $lnot C$ --- assumed [b]
5) $lnot lnot C$ --- from 3) and 4) by $lnot$elim followed by $lnot$-intro (also called Double Negation introduction), discharging [b]
6) $lnot lnot B$ --- from 5) and 2) by MT
7) $lnot A$ --- from 6) and 1) by MT
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
Note : without MT the proof is quite simple :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $lnot B$ --- from 4) and 1) by $to$-elim
6) $lnot C$ --- from 5) and 2) by $to$-elim
7) $lnot A$ --- from 4) with 3) and 6) by $lnot$-elim followed by $lnot$-intro, discharging [b]
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
add a comment |
$begingroup$
Assuming the availability of Modus Tollens rule, we have :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $lnot C$ --- assumed [b]
5) $lnot lnot C$ --- from 3) and 4) by $lnot$elim followed by $lnot$-intro (also called Double Negation introduction), discharging [b]
6) $lnot lnot B$ --- from 5) and 2) by MT
7) $lnot A$ --- from 6) and 1) by MT
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
Note : without MT the proof is quite simple :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $lnot B$ --- from 4) and 1) by $to$-elim
6) $lnot C$ --- from 5) and 2) by $to$-elim
7) $lnot A$ --- from 4) with 3) and 6) by $lnot$-elim followed by $lnot$-intro, discharging [b]
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
add a comment |
$begingroup$
Assuming the availability of Modus Tollens rule, we have :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $lnot C$ --- assumed [b]
5) $lnot lnot C$ --- from 3) and 4) by $lnot$elim followed by $lnot$-intro (also called Double Negation introduction), discharging [b]
6) $lnot lnot B$ --- from 5) and 2) by MT
7) $lnot A$ --- from 6) and 1) by MT
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
Note : without MT the proof is quite simple :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $lnot B$ --- from 4) and 1) by $to$-elim
6) $lnot C$ --- from 5) and 2) by $to$-elim
7) $lnot A$ --- from 4) with 3) and 6) by $lnot$-elim followed by $lnot$-intro, discharging [b]
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
$endgroup$
Assuming the availability of Modus Tollens rule, we have :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $lnot C$ --- assumed [b]
5) $lnot lnot C$ --- from 3) and 4) by $lnot$elim followed by $lnot$-intro (also called Double Negation introduction), discharging [b]
6) $lnot lnot B$ --- from 5) and 2) by MT
7) $lnot A$ --- from 6) and 1) by MT
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
Note : without MT the proof is quite simple :
1) $A to lnot B$ --- premise
2) $lnot B to lnot C$ --- premise
3) $C$ --- assumed [a]
4) $A$ --- assumed [b]
5) $lnot B$ --- from 4) and 1) by $to$-elim
6) $lnot C$ --- from 5) and 2) by $to$-elim
7) $lnot A$ --- from 4) with 3) and 6) by $lnot$-elim followed by $lnot$-intro, discharging [b]
8) $C to lnot A$ --- from 3) and 7) by $to$-intro, discharging [a].
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
edited Jan 17 at 13:05
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
answered Jan 17 at 11:25
Mauro ALLEGRANZAMauro ALLEGRANZA
66.5k449115
66.5k449115
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
add a comment |
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
In steps 5) and 6) you seem to apply MT for the contrapositives of 2) and 1), i.e. $lnotlnot Ctolnotlnot B$ and $lnotlnot Btolnot A$. That's a crucial point for the OP, I think..
$endgroup$
– Berci
Jan 17 at 11:30
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
$begingroup$
thank you very much for your help however my textbook does not mention double negation introduction, only elimination, and so I don't think I'm allowed to use that rule. Is there a way to prove it without DNI?
$endgroup$
– ashalrik
Jan 17 at 11:44
1
1
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
Sorry, I thought your note meant 'without MP' as in 'without modus ponens' - I thought MP was used to eliminate the conditional?
$endgroup$
– ashalrik
Jan 17 at 11:50
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
@MauroALLEGRANZA where does it say that? I can't see it anywhere. this is the textbook: people.ds.cam.ac.uk/tecb2/forallxcam.pdf however nowhere can I find DN
$endgroup$
– ashalrik
Jan 17 at 11:55
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
$begingroup$
Thank you so much for your help. I'm not familiar with Reductio however am i right in thinking this means i can introduce a contradiction then discharge (b) and use negation introduction to get to ~A?
$endgroup$
– ashalrik
Jan 17 at 12:16
add a comment |
$begingroup$
If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.
Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$displaystyle Pto Q$", and "$displaystyle Q to R$", appear on lines of a proof, "$displaystyle Pto R$" can be placed on a subsequent line.
In your argument, we then have
$(1) ;;Ato lnot B,qquad$ (premise)
$(2) ;;lnot B to lnot C,,quad$ (premise)
$(3) ;;Ato lnot C,qquad$ HS, $(1), (2)$
$(4) ;;C to lnot A,qquad$ contraposive of $(3).$
Note that instead of relying on the equivalence of an implication $pto q$ with its contrapositive, $lnot q to lnot p$, we can proceed from $(3)$ to
$(4) ;;quad C quad$ assumption
$(5) ;;quad lnotlnot Cquad$ double negation on $(4)$
$(6) ;;quad lnot Aquad$ ($3$, $5$, modus tollens)
$(7) ;; Cto lnot A quad$ ($4-6$)
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.
Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$displaystyle Pto Q$", and "$displaystyle Q to R$", appear on lines of a proof, "$displaystyle Pto R$" can be placed on a subsequent line.
In your argument, we then have
$(1) ;;Ato lnot B,qquad$ (premise)
$(2) ;;lnot B to lnot C,,quad$ (premise)
$(3) ;;Ato lnot C,qquad$ HS, $(1), (2)$
$(4) ;;C to lnot A,qquad$ contraposive of $(3).$
Note that instead of relying on the equivalence of an implication $pto q$ with its contrapositive, $lnot q to lnot p$, we can proceed from $(3)$ to
$(4) ;;quad C quad$ assumption
$(5) ;;quad lnotlnot Cquad$ double negation on $(4)$
$(6) ;;quad lnot Aquad$ ($3$, $5$, modus tollens)
$(7) ;; Cto lnot A quad$ ($4-6$)
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.
Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$displaystyle Pto Q$", and "$displaystyle Q to R$", appear on lines of a proof, "$displaystyle Pto R$" can be placed on a subsequent line.
In your argument, we then have
$(1) ;;Ato lnot B,qquad$ (premise)
$(2) ;;lnot B to lnot C,,quad$ (premise)
$(3) ;;Ato lnot C,qquad$ HS, $(1), (2)$
$(4) ;;C to lnot A,qquad$ contraposive of $(3).$
Note that instead of relying on the equivalence of an implication $pto q$ with its contrapositive, $lnot q to lnot p$, we can proceed from $(3)$ to
$(4) ;;quad C quad$ assumption
$(5) ;;quad lnotlnot Cquad$ double negation on $(4)$
$(6) ;;quad lnot Aquad$ ($3$, $5$, modus tollens)
$(7) ;; Cto lnot A quad$ ($4-6$)
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
$endgroup$
If you have Hypothetical Syllogism in your tool box, the proof becomes far more direct.
Hypothetical Syllogism (HS) is a valid argument form in which two premises are conditionals, and the rule is that, whenever instances of "$displaystyle Pto Q$", and "$displaystyle Q to R$", appear on lines of a proof, "$displaystyle Pto R$" can be placed on a subsequent line.
In your argument, we then have
$(1) ;;Ato lnot B,qquad$ (premise)
$(2) ;;lnot B to lnot C,,quad$ (premise)
$(3) ;;Ato lnot C,qquad$ HS, $(1), (2)$
$(4) ;;C to lnot A,qquad$ contraposive of $(3).$
Note that instead of relying on the equivalence of an implication $pto q$ with its contrapositive, $lnot q to lnot p$, we can proceed from $(3)$ to
$(4) ;;quad C quad$ assumption
$(5) ;;quad lnotlnot Cquad$ double negation on $(4)$
$(6) ;;quad lnot Aquad$ ($3$, $5$, modus tollens)
$(7) ;; Cto lnot A quad$ ($4-6$)
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
edited Jan 17 at 20:48
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
answered Jan 17 at 12:57
jordan_glenjordan_glen
1
1
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076856%2fnatural-deduction-logic-proof-help%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
What is MT? Which inference rules are you allowed to use?
$endgroup$
– Taroccoesbrocco
Jan 17 at 11:22
$begingroup$
MT is modus tollens. All inference rules should be permitted, I'm learning from the forallx Cambridge textbook.
$endgroup$
– ashalrik
Jan 17 at 11:29