______________________________

[[Example 9.1.1]]

Consider the following arguments:

A É B

A

\ B

 

(A · B) É (D v C)

A · B

\ (D v C)

Both of these arguments are substitution instances of the elementary valid argument form modus ponens:

p É q

p

\ q

__________________________________

[[Example 9.2.1]]

Consider this argument:

(A v B) É (~ ~D · C)

A

\ D

One formal proof of this argument is the following:

    1. (A v B) É (~ ~D · C)
    2. A

      3. \ D

    3. A v B 2, Add.
    4. ~ ~D · C1,3, M.P.
    5. ~ ~D4, Simp.
    6. D5, D.N.

Another equally valid proof is this:

    1. (A v B) É (~ ~D · C)
    2. A

      3. \ D

    3. (A v B) É (D · C)1, D.N.
    4. A v B 2, Add.
    5. D · C3,4, M.P.
    6. D4, Simp.

In both cases we replaced the expression "~ ~D" with the logically equivalent expression "D."

 

__________________________________

[[Example 9.2.1]]

Consider the following argument:

If the stock market continues to fall, then Oliver will have to get another job.

If Oliver looses his financial optimism, he will have to get another job.

Therefore if the stock market continues to fall, then Oliver will loose his financial optimism.

The argument can be symbolized as follows:

S É J

L É J

\ S É L

 

In order to prove that this argument is invalid, we have to find an assignment of truth values for the propositions S, J, and L that makes the conclusion false and both premisses true. Let us start with the conclusion S É L. In order to make this conditional conclusion false, S must be true and L false. If we now assume that J is true, then the premisses S É J and L É J are both true.

S

J

L

S É J

L É J

S É L

T

T

F

T

T

F

This shows that there exists an assignment of truth values that makes all the premisses true and the conclusion false. It follows therefore that the argument is invalid.

__________________________

[[Example 9.4.1]]

 

    1. P · ~P

      2. \ G

    2. P1, Simp
    3. ~P · P1, Com.
    4. ~ P3, Simp
    5. P v G2, Add
    6. G 5, 4, D.S.