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1 Discrete Structures Lecture 12 Implication III Read: Ch 4.1.

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Presentation on theme: "1 Discrete Structures Lecture 12 Implication III Read: Ch 4.1."— Presentation transcript:

1 1 Discrete Structures Lecture 12 Implication III Read: Ch 4.1

2 2 Additional Theorems re:  (4.1) p  (q  p) (4.2) Monotonicity of V: proof in text (p  q)  (p V r  q V r) (4.3) Monotonicity of  : proof in class (p  q)  (p  r  q  r)

3 3 Abbreviation for Proving  The transitivity of  allows us to use an abbreviation for proofs involving implication. (3.82) Transitivity: (a) (p  q)  (q  r)  (p  r) P  Q R By transitivity (3.82 a), this proves that P  R

4 4 Combining  and  The mutual transitivity of  and  allows us to use a combination of Leibniz steps and the abbreviation using . (3.82) Transitivity: (b) (p  q)  (q  r)  (p  r) P  Q  R By transitivity (3.82b), this proves that P  R.

5 5 Combining  and  The mutual transitivity of  and  allows us to use a combination of Leibniz steps and the abbreviation using . (3.82) Transitivity: (c) (p  q)  (q  r)  (p  r) P  Q  R By transitivity (3.82c), this proves that P  R.

6 6 Using the Correct Hints When using this abbreviation, one must be careful to use the correct hints. Example of an incorrect hint: Assuming P1  Q1 is a theorem, we have x  P1  x  Q1 The correct hint SHOULD be: x  P1  x  Q1 Incorrect!

7 7 Abbreviation for Proving  cont. However, before this can be used, R must first be proved. Here is one proof of R (assuming P1  Q1 is a theorem). (P1  Q1)  (x  P1  x  Q1) = true  (x  P1  x  Q1) = x  P1  x  Q1

8 8 Abbreviation for Proving  cont. This is tortuous, so we can abbreviate as follows. x  P1  x  Q1

9 9 (4.3) Monotonicity of  : (p  q)  (p  r  q  r) Problem 4.2 says to prove (4.3) p  r  q  r = ¬(p  r) V (q  r) = ¬p V ¬r V (q  r) = (¬p V ¬r V q)  (¬p V ¬r V r) = ¬p V ¬r V q  ¬p V q = p  q

10 10 A Diversion on Abbreviation Mistakes There may be a tendency to try to write E[z:= P1]  E[z:= Q1]

11 11 Diversion  Continued The step on the previous slide is not always sound. Here is an unsound use of this technique. ¬P1  ¬Q1 Here is a sound use. ¬P1  ¬Q1

12 12 Metatheorem Parity This metatheorem explains exactly how a replacement of P by Q is to be made, where P  Q.

13 13 Metatheorem Parity Continued Metatheorem Parity. Consider a boolean expression E that: contains only operators ¬, V, and  and has a single occurrence of a variable z. Suppose P  Q is a theorem.

14 14 Metatheorem Parity If z occurs within the scope of an even number of ¬ operations, then E[z:= P]  E[z:= Q]. If z occurs within the scope of an odd number of ¬ operations, then E[z:= P]  E[z:= Q].

15 15 Metatheorem Parity For example, given P  Q, Metatheorem Parity implies the following. R  P  R  Q R  ¬P  R  ¬Q

16 16 Metatheorem Parity For example, given P  Q, Metatheorem Parity implies the following. ¬(R  P)  ¬(R  Q) ¬(R  ¬P)  ¬(R  ¬Q)

17 17 Metatheorem Parity For example, given P  Q, Metatheorem Parity implies the following. ¬(R  P)  ¬(R  Q) ¬R V ¬P  ¬R V ¬Q ¬(R  ¬P)  ¬(R  ¬Q) ¬R V ¬¬P  ¬R V ¬¬Q Still even for Q’s scope. In P/Q’s scope, only one negative.

18 18 Bonus Problem (4.4 in book) (p  q)  (r  s)  (p V r  q V s) Problem 4.4 says to prove the above theorem (p  q)  (r  s)


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