Constructive dilemma (CD)

Given a conjunction of two conditionals and a disjunction, one disjunct of which is the antecedent of one of the conjoined conditionals and the other the antecedent of the other conjoined conditional, you may conclude that a disjunction of the two consequents is true.

Given: p v r, (p > q) * (r v s)

Conclude: q v s

Simplification (SIMP)

Given a conjunction you may conclude the first conjunct is true.

Given: p * q

Conclude: p

Conjunction (CONJ)

Given any two statements you may conclude that the conjunction of those two statements is true.

Given: p, q

Conclude: p * q

Addition (ADD)

Given any statement you may conclude that that statement disjoined with any other is true.

Given: p

Conclude: p v q

Example 1:

1. A > B

2. A * C

3. (B v C) > D /D

Strategy: Looking at line 3 you see that D is the consequent of a conditional and could be detached using MP should you prove that B v C. B is the consequent of the conditional in line 1. If you could detach B you could ADD any disjunct, in particular you could ADD C to prove the antecedent for line 3. To detach B from line 1 you need to show that A (the antecedent of 1) is true. Line 2 says that both A and C are true. You can simplify 2 to get A alone.

4. A ___2, SIMP

5. B ___1, 4, MP

6. B v C ___5, ADD

7. D ___3, 6, MP

Example 2:

1. (M > N) * (~O > P)

2. M ___/N v P

Strategy: The conclusion is the disjunction of the two conditionals that are conjoined in 1. That suggests the use of CD. But you need the disjunction of the two antecedents. In line 2 you are given M. You can ADD ~O to M.

3. M v ~O ___2, ADD

4. N v P ___1, 3, CD

Example 3:

1. D v E

2. (F * E) > G

3. ~D

4. F ___/G

Strategy: To detach G from line 2 you know that both F and E are true. If you could show that E is true you could conjoin (CONJ) it with F, which you are given in line 4. E can be detached from line 1 if you can show that D is false. Line 3 gives you ~D.

5. E ___1, 3, DS

6. F * E ___4, 5, CONJ

7. G ___2, 6, MP

Why are the following mistakes?

1. A > (B > C)

2. B

3. C ___1, 2, MP

1. A v (B * C)

2. B ___1, SIMP

1. A

2. A * B ___1, ADD

1. A v B

2. A ___1, SIMP

1. A > B

2. A > (B v C) ___1, ADD

1. (A > B) > C

2. ~ B

3. ~ A ___1, 2, MT

1. A > B

2. A > C

3. B > C ___1, 2, HS

1. A > B

2. A > C

3. B * C ___1, 2, CONJ

1. ~(A * B)

2. ~A ___1, SIMP

1. ~(A v B)

2. ~ A

3. B ___1, 2, DS