Checking arguments for validity

There are two methods for checking the validity of an argument, the direct and the indirect method.

Direct Method

Suppose you have an argument A with premise P₁,...P_n, and conclusion C.  Then one tries to derive the conclusion from the premises by reasoning correctly. To keep the method informal, I provide no rules that must be followed in the derivation; however, the following are useful:
I.
A->B
A
------      (This is Modus Ponens, MP)
B

II.
A->B
-B
-----      (This is Modus Tollens, MT)
-A

III.
AvB
-A
-------
B

IV.
-(A&B)
A
--------
-B

V.
A<->B
-A
--------
-B

VI.
-(AvB)
---------
-A&-B

VII.
-(A->B)
----------
A&-B

VIII.
A->B
B->C
--------
A->C

Exercise: Check by truth table that (I-VIII) are valid.

Example

Let's check the following argument for validity:
-A v B
-B
C->A
-----
-C

First, let's list the premises, and then try and derive (validly) the conclusion:
1. -(A& -B) (Premise)
2. -B (Premise)
3. C->A (Premise)
4. -A (From 1, 2 by III)
5. -C        (From 3, 4, by MT)

Since we obtained the conclusion from the premises by reasoning correctly, the argument is valid. Note that:

1. each step is numbered
2. each step is justified by showing from which lines it is derived.  (Here I also state the rule used, but you needn't do this).

Sometimes the direct method does not work well. Then we should attempt the indirect method, which is sure to work but is a bit more complicated than the direct.

Indirect Method

Suppose you have an argument A with premise P₁,...P_n, and conclusion C. Then, combine P₁,...P_n, and -C, and try to derive a contradiction, i.e., a statement of the form Q& -Q. Of course, the derivation must be valid, i.e., you must reason correctly in it.   If you derive a contradiction, then you have proved that A is valid.  (WHY?  Think about it, before you download this)

Now, let's check the following argument for validity.

M->N
-P-> -(N&O)
----------
O->(N->P)

A brief look convinces us that the direct method will not really work. So, let’s use the indirect method

1.     M->N
2.    -P-> -(N&O)
3.    -[O->(N->P)]
4.    O & -(N->P)        (From 3, by VII)
5.    O                        (From 4)
6.    -(N->P)               (From 4)
7.    N& -P                 (from 6, by VII)
8. N                        (From 7)
9.    -P                       (From 7)
10.   -(N&O)              (From 2, 9, by MP)
11.   -O                      (From 8, 10, by IV)
12.   O& -O                (From 5, 11)

Line 12 is a contradiction. Hence, the argument is valid.