How to prove that “n number of proposition(s) give rise to 2^(n) number of truth-value combinations”...
$begingroup$
In proposition logic, alphabets are used to represent atomic propositions, understood as a grammatically correct expression in formal language.
Every atomic proposition is either true or false and the combination of atomic propositions using logical connectives (such as and/or/if,then/iff) give rise to complex propositions.
For instance, the complex proposition (A & B) is true if and only if both atomic propositions A and B are true. In other instances, where either A or B or both are false, the complex proposition is false.
When trying to evaluate the truth conditions for complex propositions with many atomic propositions, one's required to compute the possible truth-value combinations among the different atomic propositions first. For instance, to evaluate the truth-conditions for the proposition (A&B), one needs to list the possible truth-value combinations as (A true B true, A false B False, A True B False, A False B True) first.
With this in mind, how do I prove, using mathematical induction, that the number of possible truth-value combinations for n propositions is 2^n?
I am not exposed to set theory and it is my humble request that any explanations involving set theory be as beginner-friendly as possible.
yt
logic
asked Dec 20 '18 at 11:02
user628101user628101
132
$endgroup$
add a comment |
$begingroup$
In proposition logic, alphabets are used to represent atomic propositions, understood as a grammatically correct expression in formal language.
Every atomic proposition is either true or false and the combination of atomic propositions using logical connectives (such as and/or/if,then/iff) give rise to complex propositions.
For instance, the complex proposition (A & B) is true if and only if both atomic propositions A and B are true. In other instances, where either A or B or both are false, the complex proposition is false.
When trying to evaluate the truth conditions for complex propositions with many atomic propositions, one's required to compute the possible truth-value combinations among the different atomic propositions first. For instance, to evaluate the truth-conditions for the proposition (A&B), one needs to list the possible truth-value combinations as (A true B true, A false B False, A True B False, A False B True) first.
With this in mind, how do I prove, using mathematical induction, that the number of possible truth-value combinations for n propositions is 2^n?
I am not exposed to set theory and it is my humble request that any explanations involving set theory be as beginner-friendly as possible.
yt
logic
asked Dec 20 '18 at 11:02
user628101user628101
132
$endgroup$
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59
add a comment |
$begingroup$
In proposition logic, alphabets are used to represent atomic propositions, understood as a grammatically correct expression in formal language.
Every atomic proposition is either true or false and the combination of atomic propositions using logical connectives (such as and/or/if,then/iff) give rise to complex propositions.
For instance, the complex proposition (A & B) is true if and only if both atomic propositions A and B are true. In other instances, where either A or B or both are false, the complex proposition is false.
When trying to evaluate the truth conditions for complex propositions with many atomic propositions, one's required to compute the possible truth-value combinations among the different atomic propositions first. For instance, to evaluate the truth-conditions for the proposition (A&B), one needs to list the possible truth-value combinations as (A true B true, A false B False, A True B False, A False B True) first.
With this in mind, how do I prove, using mathematical induction, that the number of possible truth-value combinations for n propositions is 2^n?
I am not exposed to set theory and it is my humble request that any explanations involving set theory be as beginner-friendly as possible.
yt
logic
asked Dec 20 '18 at 11:02
user628101user628101
132
$endgroup$
In proposition logic, alphabets are used to represent atomic propositions, understood as a grammatically correct expression in formal language.
Every atomic proposition is either true or false and the combination of atomic propositions using logical connectives (such as and/or/if,then/iff) give rise to complex propositions.
For instance, the complex proposition (A & B) is true if and only if both atomic propositions A and B are true. In other instances, where either A or B or both are false, the complex proposition is false.
When trying to evaluate the truth conditions for complex propositions with many atomic propositions, one's required to compute the possible truth-value combinations among the different atomic propositions first. For instance, to evaluate the truth-conditions for the proposition (A&B), one needs to list the possible truth-value combinations as (A true B true, A false B False, A True B False, A False B True) first.
With this in mind, how do I prove, using mathematical induction, that the number of possible truth-value combinations for n propositions is 2^n?
I am not exposed to set theory and it is my humble request that any explanations involving set theory be as beginner-friendly as possible.
yt
logic
logic
asked Dec 20 '18 at 11:02
user628101user628101
132
asked Dec 20 '18 at 11:02
user628101user628101
132
asked Dec 20 '18 at 11:02
user628101user628101
132
asked Dec 20 '18 at 11:02
user628101user628101
132
asked Dec 20 '18 at 11:02
user628101user628101
132
132
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59
add a comment |
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is how to do with mathematical induction:
when $n = 1$ then $2^n = 2$. Which is true since you have two truth values. Hence, the base case is true.
For the inductive step. We want to show that $n+1$ is true whenever $n$ is true. But $2^{n+1} = 2^ntimes 2$ which is the number of truth values for the proposition $n+1$ multiplied by the number of truth values of the previous $n$ propositions which sums up to the truth values of all propositions up to $n+1$.
We can grasp it as follows with the analogy of coins. Suppose we have 1 coin. Now, there are two truth values for this coin. If we added another new coin, then we will have two truth values for this new coin. The combination of the truth values of this new coin and the old one is $2*2=2^2=4$. Similarly if we have added third new coin, we will have $2^2 * 2 = 2^3 = 8$ combinations for these three coins and so on. I hope the idea is getting clear now.
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
$endgroup$
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
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%2f3047418%2fhow-to-prove-that-n-number-of-propositions-give-rise-to-2n-number-of-truth%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is how to do with mathematical induction:
when $n = 1$ then $2^n = 2$. Which is true since you have two truth values. Hence, the base case is true.
For the inductive step. We want to show that $n+1$ is true whenever $n$ is true. But $2^{n+1} = 2^ntimes 2$ which is the number of truth values for the proposition $n+1$ multiplied by the number of truth values of the previous $n$ propositions which sums up to the truth values of all propositions up to $n+1$.
We can grasp it as follows with the analogy of coins. Suppose we have 1 coin. Now, there are two truth values for this coin. If we added another new coin, then we will have two truth values for this new coin. The combination of the truth values of this new coin and the old one is $2*2=2^2=4$. Similarly if we have added third new coin, we will have $2^2 * 2 = 2^3 = 8$ combinations for these three coins and so on. I hope the idea is getting clear now.
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
$endgroup$
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
add a comment |
$begingroup$
Here is how to do with mathematical induction:
when $n = 1$ then $2^n = 2$. Which is true since you have two truth values. Hence, the base case is true.
For the inductive step. We want to show that $n+1$ is true whenever $n$ is true. But $2^{n+1} = 2^ntimes 2$ which is the number of truth values for the proposition $n+1$ multiplied by the number of truth values of the previous $n$ propositions which sums up to the truth values of all propositions up to $n+1$.
We can grasp it as follows with the analogy of coins. Suppose we have 1 coin. Now, there are two truth values for this coin. If we added another new coin, then we will have two truth values for this new coin. The combination of the truth values of this new coin and the old one is $2*2=2^2=4$. Similarly if we have added third new coin, we will have $2^2 * 2 = 2^3 = 8$ combinations for these three coins and so on. I hope the idea is getting clear now.
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
$endgroup$
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
add a comment |
$begingroup$
Here is how to do with mathematical induction:
when $n = 1$ then $2^n = 2$. Which is true since you have two truth values. Hence, the base case is true.
For the inductive step. We want to show that $n+1$ is true whenever $n$ is true. But $2^{n+1} = 2^ntimes 2$ which is the number of truth values for the proposition $n+1$ multiplied by the number of truth values of the previous $n$ propositions which sums up to the truth values of all propositions up to $n+1$.
We can grasp it as follows with the analogy of coins. Suppose we have 1 coin. Now, there are two truth values for this coin. If we added another new coin, then we will have two truth values for this new coin. The combination of the truth values of this new coin and the old one is $2*2=2^2=4$. Similarly if we have added third new coin, we will have $2^2 * 2 = 2^3 = 8$ combinations for these three coins and so on. I hope the idea is getting clear now.
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
$endgroup$
Here is how to do with mathematical induction:
when $n = 1$ then $2^n = 2$. Which is true since you have two truth values. Hence, the base case is true.
For the inductive step. We want to show that $n+1$ is true whenever $n$ is true. But $2^{n+1} = 2^ntimes 2$ which is the number of truth values for the proposition $n+1$ multiplied by the number of truth values of the previous $n$ propositions which sums up to the truth values of all propositions up to $n+1$.
We can grasp it as follows with the analogy of coins. Suppose we have 1 coin. Now, there are two truth values for this coin. If we added another new coin, then we will have two truth values for this new coin. The combination of the truth values of this new coin and the old one is $2*2=2^2=4$. Similarly if we have added third new coin, we will have $2^2 * 2 = 2^3 = 8$ combinations for these three coins and so on. I hope the idea is getting clear now.
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
edited Dec 20 '18 at 11:44
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
answered Dec 20 '18 at 11:18
Maged SaeedMaged Saeed
8671417
8671417
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
add a comment |
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
1
1
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hi! Thank you for your reply. I'm actually interested in finding out how to prove that the number of possible combinations is 2^n. A watered-down version of the problem is this. Suppose that a coin either lands on heads or tails and there are n coins, how do we prove that the total number of combinations of heads and tails is 2^n?
$endgroup$
– user628101
Dec 20 '18 at 11:29
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101 if you are looking for mathematical induction. Then I guess I did so.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:36
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
$begingroup$
Hey @user628101. I have updated the answer accordingly. Do you still have any problems? Please reply.
$endgroup$
– Maged Saeed
Dec 20 '18 at 11:45
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%2f3047418%2fhow-to-prove-that-n-number-of-propositions-give-rise-to-2n-number-of-truth%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
$begingroup$
I think you mean $2^{2^n}$. When $n=1$ there are four functions $f(x)$: (i) $f(x)=T$ for all $x$; (ii) $f(x)=F$ for all $x$; (iii) $f(x)=x$ for all $x$; (iv) $f(x)=not-x$ for all $x$.
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:12
$begingroup$
Certainly not! When there are 3 atomic propositions, the combinations are: TTT, FFF, TTF, TFT, FTT, FFT, TFF, FTF, which is 2^3 = 8 propositions.
$endgroup$
– user628101
Dec 20 '18 at 11:18
$begingroup$
I am sorry, I thought you wanted the number of what you were calling "complex propositions".
$endgroup$
– ancientmathematician
Dec 20 '18 at 11:59