Well I'm sorry, but really... I'm refuting this: If I can style, I am a typesetter.

Given the following premises:

Typesetter is someone who can typeset. If one can typeset, one can style text and one can do graphics too.

Is the following conclusion valid?

If I can style, I am a typesetter.

Let T = typesetting

Let S = Text Styling

Let G = Graphics

Given:

T -> SG

Prove:

S->T is valid.

Use the Truth Tables:

T | S | G | T->SG | S->T | (S->T)->(T->SG) (Conclusion) |

0 | 0 | 0 | 1 | 1 |

0 | 0 | 1 | 1 | 1 |

0 | 1 | 0 | 1 | 0 |

0 | 1 | 1 | 1 | 0 |

1 | 0 | 0 | 0 | 1 | * (I can typeset, I can't style, nor do I know how to do graphics [FAIL]) |

1 | 0 | 1 | 0 | 1 | * (I can typeset, I can't style, but I know how to do graphics [FAIL]) |

1 | 1 | 0 | 0 | 1 | * (I can typeset, I know how to style, but I don't know how to do graphics [FAIL]) |

1 | 1 | 1 | 1 | 1 |

I haven't done truth table in a while, so please be considerate, but if you don't know what is it, go study the following links:

http://en.wikipedia.org/wiki/Truth_table
http://en.wikipedia.org/wiki/Boolean_algebra
Let's represent 1 = true, and 0 = false. The * in the last column are the ones that fail that clause (whether styling concludes to be a typesetting)

As one can see, that conclusion is invalid. If you can refute my logic this way, please say so but in a boolean algebra or truth table fashion rather than words. Basic boolean stuff.

Even without "G", it's still invalid, as T->S =/=> S->T... Basic Definition (note: T->S == !T + S). I just add in "G" for a bit more meanings.