How to prove that “n number of proposition(s) give rise to 2^(n) number of truth-value combinations”...












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
















0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 11:44

























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
















  • 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




















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