Reductio Rule (RA)

Agenda

1. Quiz
2. Questions about HW?
3. New Rule: RA
4. Proofs

Quiz

A. 

1. ~A&B
2. (~Av(B>~C))>(Dv~B)   /:. D

B. 

1. (Pv~Q)>R
2. P&S  /:.  ~(S>~R)

C. 

1. If you have only TWENTY % in the course you will GET an A only if you CHALLENGE and DEFEAT me in a judo death match, but if you WORK really hard you can still PASS.
2. I'm usually GRUMPY unless someone brings me either CHOCOLATE chip or PEANUT butter cookies.

Reductio (RA)

Solve:

1. ~Q&P  /:.  ~(P>Q)

How to set up an RA proof: Suppose I need to solve for ~(P>Q)

Step 1. Do all the top down rules you can.

Step 2. RA is a bottom up rule. We look at the last line of our proof that doesn't have a justification and we set up from there. Leave space and write the 'opposite' of the line you're trying to solve.

1.  ~Q&P    A

2. ~Q          &E 1

3. P             &E 1

4. P>Q      ARA

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n. ~(P>Q)  RA

Step 3. Try to derive a contradiction by putting two 'opposites' together with &I

1.  ~Q&P    A

2. ~Q          &E 1

3. P             &E 1

4. P>Q       ARA

5. Q           MP 3, 4

6. Q&~Q   &I 2, 5

7. ~(P>Q)  RA 4, 6

RA Proofs
A.

1.  T>P
2.  (P&~Q)>~(S&~Q)  /:. (P>~Q)>~(T&S)

B.

1. ~P
2. (Pv~Q)>Q  /:. ~(Pv~Q)

C. 

1. D<>~(A&~B)
2. ~(D&C)v(A&~B)  /:.  ~(A&~B)>~(D>C)

D. 

1. ~D>(A&C)
2. (B&D)>E
3. (DvF)>~E  /:. ~(A&C)>~(Bv~D)

E. 

1. (R&S)>~(P>~Q)
2. ~(T>P)>(R&S)
3. T
4. ((T>P)&P)>Q        /:. ~(P>Q)

F.

1. (~D&~E)>F
2. A&~D
3. ((F&~B)vG)>~(A>~B)    /:.  (A>~B)>~(Dv~E)