JUSTIFYING STEPS IN A PROOF

**A
proof is a finite series of formulas,
beginning with the premises of an argument and ending with the conclusion, in
which each line following the premises is justified according to accepted
rules of inference and
equivalence. A proof is a formal demonstration that the conclusion of the
argument follows from the premises. Any argument for which a proof can be
constructed is valid, and for any valid argument a proof can be constructed.**

Each line in a proof must
be * justified* in some way, that is, some reason must be given for
believing it to be true. Remember, validity preserves truth. Each line in an
argument is either a premise or a

Let's see how this works. Consider the argument:

[~(P
●
Q)
→
(~R
→
~S)], ~(P ●
Q)
∴(~R
→
~S) |

First we need to set forth our premises, numbering each line so that we can easily refer back to it, and then justifying the premises by identifying them as premises.

Line # | Argument (instance) | Argument Form | Justification |

1 | [~(P ● Q) → (~R → ~S)] | premise | |

2 | ~(P ● Q) | premise |

Now, Identify an argument form or inference rule that maps onto your premises. In an actual proof we don' have a column for the argument form, but I have included it here to help you see just what is going on in the justification.

Line # | Argument (instance) | Argument Form | Justification |

1 | [~(P ● Q) → (~R → ~S)] | (p → q) | premise |

2 | ~(P ● Q) | p | premise |

Finally, create a new line, a substitution instance of the conclusion of the argument form using the proper WFF's to replace the sentential variables in the form.

Line # | Argument (instance) | Argument Form | Justification |

1 | [~(P ● Q) → (~R → ~S)] | (p → q) | premise |

2 | ~(P ● Q) | p | premise |

3 |
∴(~R
→
~S) |
∴q |
1,2 Modus Ponens |

And that's all there is to it.

Here is a slightly more complicated example, in which I leave out the argument forms:

A, B,
(A
●
B)
→** **
C
**
∴**C

1 | A | premise |

2 | B | premise |

3 | (A · B) → C | premise |

4 | (A · B) | 1,2 conjunction |

5 | C | 3,4 modus ponens |

Finally, a more complicated example, consider the argument:

**(A▼B)**→**C **
**
****
∴**** A**→**C**

Line Inf/Equiv Rule applied Justification 1 (A ▼B)→Cpremise 2 ~(A ▼B)▼C(p → q) :: (~p ▼q) 1, implication 3 (~A ● ~B) ▼C~(p ● q) :: (~p ▼ ~q) 2, DeMorgan 4 C ▼(~A ●· ~B)(p ▼ q) :: (q ▼ p) 3, commutation 5 (C ▼~A) ● (C▼~B)p▼(q ● r) :: (p ▼q) ● (p▼ r) 4, distribution 6 C ▼~A(p ● q) ∴p5, simplification 7 ~A ▼C(p ▼ q) :: (q ▼ p) 6, commutation 8 A→C (p → q) :: (~p ▼q) 7, implication

Note that this rather complicated argument uses only 1 inference rule, and that a very simple one (simplification). All of the heavy lifting in this argument is done by the equivalence rules.