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Given that the set {∨, $\wedge$ , ¬} is functionally complete, how would I prove whether the set $\{\to\}$ is functionally complete?

expressing $→$ in terms of $∨$: $¬A∨B$

expressing $→$ in terms of $∧$: $¬(A∧¬B)$

I understand the above two expressions, but cannot seem to prove that it is/is not functionally complete.

bawse
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user141834
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  • Your examples go the wrong way: they express $\implies$ in terms of other operations. To show ${\implies}$ is functionally complete, you would need to express the other operations in terms of $\implies$. – Noah Schweber Oct 16 '15 at 03:55
  • You say "expressing $\rightarrow$ in terms of $\vee$" (which can't be done) but then you express $\rightarrow$ in terms of $\vee$ and $\neg$. So, when you say "is ${\rightarrow}$ functionally complete", do you mean "is ${\rightarrow}$ funcitonally complete", or do you mean "is ${\rightarrow,\neg}$ functionally comnplete"? – bof Oct 16 '15 at 04:02
  • I meant is {→} functionally complete. I made a mistake – user141834 Oct 16 '15 at 04:04

1 Answers1

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Suppose I have an expression $\pi$ involving propositional variables $A_1, . . . , A_n$, but only the connective "$\implies$" (OK, fine, and parentheses).

  • What can I say about the truth value of $\pi$ if I assign "True" to each of the $A_1, . . . , A_n$?

  • What does that tell you about whether $\{\implies\}$ is functionally complete?

Noah Schweber
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    The answer to your first bullet point: Would the truth value of $\pi$ be true? – user141834 Oct 16 '15 at 04:00
  • Seems plausible - can you prove it? (The answer is yes, you're right - but this technically requires a proof.) – Noah Schweber Oct 16 '15 at 04:01
  • I have no idea of where to even start with the proof for that...the way in which I answered your question was just based off the definition of the operator itself – user141834 Oct 16 '15 at 04:02
  • @user141834 also in the same boat, looking at a similar question. Not sure how to proceed – bawse Oct 16 '15 at 04:10
  • Well, let's start at the big picture: based on your guess - which is correct, but not yet proved - do you think ${\implies}$ is functionally complete? If so: why? If not: is there some expression you don't think you can use $\implies$ alone to define? – Noah Schweber Oct 16 '15 at 04:26
  • I don't think $\to$ is functionally complete, however that is a complete guess. With that being said, I don't think $\to$ can be used to define the $\wedge$ operator, simply because both have different truth tables.. – user141834 Oct 16 '15 at 04:31
  • @user141834 While it's correct that $\implies$ can't be used to define $\wedge$, that uses a different argument than the one which my hint above started. What's an expression that isn't made true if each of its components is true? – Noah Schweber Oct 16 '15 at 04:32
  • Alternatively, if you want to show that $\implies$ can't be used to define $\wedge$: try the same exercise as in my answer, but this time making each $A_i$ false. What happens now? – Noah Schweber Oct 16 '15 at 04:40
  • In line with the argument that you have used, an expression that isn't made true if each of its components are true is one which uses the $\neg$ operator. Is this correct? – user141834 Oct 16 '15 at 04:55
  • @user141834 Sure - for instance, the expression "$\neg A_1$." If you believe that any expression involving only "$\implies$" (and parentheses) is made true when each of the propositional variables in it is made true, does this convince you that you can't express $\neg$ in terms of $\implies$ alone? If not: why not? If so: then you're halfway there - you just need to show that your answer to my exercise was, in fact, correct. (For that, here's a hint: induction . . .) – Noah Schweber Oct 16 '15 at 04:59
  • Ah I see. I had a feeling induction was coming, but induction on what property? – user141834 Oct 16 '15 at 05:02
  • @user141834 Basically, you always want to induct on the complexity of the thing you're looking at, informally. So: what's a good way to measure the complexity of a formula? For instance, surely we agree that "$((B\implies C)\implies(A\implies C))\implies((A\implies B)\implies C)$" is more complicated than "$A\implies B$"; why? – Noah Schweber Oct 16 '15 at 05:05
  • I am not familiar with the concept of complexity of the formula, could you explain further? – user141834 Oct 16 '15 at 05:11
  • @Noah Schweber I've been through this with userNNN already, by the way. – BrianO Oct 16 '15 at 06:18
  • I see, thanks. @user141834 the argument is exactly what BrianO did in the other question http://math.stackexchange.com/questions/1481376/proof-of-functional-completeness-of-logical-operators. – Noah Schweber Oct 16 '15 at 06:27