Simple logic truth table question

Please help with this question, my English is not that good:

Here is a truth table for the new symbol *

P | * | Q |

T | F | T |

T | T | F |

F | T | T |

F | T | F |

Given this truth-table, what ordinary English expression can this new truth-functional connective (*) be used to symbolize?

One other broad question:

Suppose there are two sentences a and b. On every interpretation that a is true, b is false. What can you conclude (if anything) about the following argument? Briefly explain.

~(a -> b)

.:. ~b v a

-> :material conditional

v :disjuction/or

Many thanks for help

Re: Simple logic truth table question

Quote:

Originally Posted by

**apterosss** Please help with this question, my English is not that good:

Here is a truth table for the new symbol *

P Q P * Q

T T F

T F T

F T T

F F T

Given this truth-table, what ordinary English expression can this new truth-functional connective (*) be used to symbolize?

I have most often have seen this as $\displaystyle \begin{array}{*{20}{c}} P&Q&\;&{P|Q} \\ \hline T&T&\;&F \\ T&F&\;&T \\ F&T&\;&T \\ F&F&\;&T \end{array}$.

WVO Quine called this the *alternative denial operator*, first discovered by CS Peirce.

These are equivalent:

$\displaystyle \begin{array}{*{20}{c}} {\neg P}&{\mathop = \limits^{df} }&{P|P} \\ {P \wedge Q}&{\mathop = \limits^{df} }&{(P|Q)|(P|Q)} \\ {P \vee Q}&{\mathop = \limits^{df} }&{(P|P)|(Q|Q)} \\ {P \to Q}&{\mathop = \limits^{df} }&{P|(Q|Q)} \end{array}$

Re: Simple logic truth table question

Plato - apologies, something happened with my spacing.

Yes your table is equivalent to what I tried to put; please see edit.

What would the English phrase be for this? Is there a sentence this could be symbolised with?

Thanks :)

Re: Simple logic truth table question

Quote:

Originally Posted by

**apterosss** Given this truth-table, what ordinary English expression can this new truth-functional connective (*) be used to symbolize?

A * B can be read, "A and B cannot both be true" or "A contradicts B" (note that this is the same as "B contradicts A", i.e., * is symmetric).

Quote:

Originally Posted by

**apterosss** Suppose there are two sentences a and b. On every interpretation that a is true, b is false. What can you conclude (if anything) about the following argument? Briefly explain.

~(a -> b)

.:. ~b v a

This argument is valid without assuming anything about truth of a and b in interpretations.

Re: Simple logic truth table question

Quote:

Originally Posted by

**apterosss** What would the English phrase be for this? Is there a sentence this could be symbolised with?

Actually there are probably many names for this.

As I said, WVO Quine called this the **alternative denial operator**r, first discovered by CS Peirce.

I don't remember is Perice had a name for it.

It also goes by the name *stroke function*,

The chart I posted shows that the other functions can be defined using this operator.

It in effect: "It is false if both statements are true and true otherwise".

Re: Simple logic truth table question

Hello, apterosss!

Quote:

Here is a truth table for the new symbol $\displaystyle \star$

. . $\displaystyle \begin{array}{ccc} P & \star & Q \\ \hline T & F & T \\ T & T & F \\ F & T & T \\ F & T & F \end{array}$

Given this truth-table, what ordinary English expression

can this new operation $\displaystyle (\star)$ be used to symbolize?

$\displaystyle \text{We have: }\:\sim(P \wedge Q) \:\Longleftrightarrow\: \sim\!P\:\vee \sim\!Q$

. . $\displaystyle "P\text{ or }Q\text{ is false."}$

Re: Simple logic truth table question

Thank you all for your help :)

Re: Simple logic truth table question

Sorry I mistyped this one - still don't understand the answer.

One other broad question:

Suppose there are two sentences a and b. On every interpretation that a is true, b is false. What can you conclude (if anything) about the following argument? Briefly explain.

~(a -> c)

.:. ~b v c

-> :material conditional

v :disjuction/or

Re: Simple logic truth table question

Quote:

Originally Posted by

**apterosss** Suppose there are two sentences a and b. On every interpretation that a is true, b is false. What can you conclude (if anything) about the following argument?

~(a -> c)

.:. ~b v c

This argument is valid in this sense. The formula ~(a -> c) is equivalent to a /\ ~c. Then use the assumption.