Proofs with Predicate Calculus ITCS 2175 Revision: 1.0. Author: Zachary Wartell Copyright Zachary Wartell, UNCC, 2010 Contributors:?

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1 Proofs with Predicate Calculus ITCS 2175 Revision: 1.0 Author: Zachary Wartell Copyright Zachary Wartell, UNCC, 2010 Contributors:? This document supplements the course packets and the textbook. 1. Definitions: AF,AT,ALOF,ALOT Definitions: AF,AT,ALOF,ALOT Syntax Definition Contrapositive all false AF P(x) x P(x) x P(x) all true AT P(x) x P(x) x P(x) at least one false ALOF P(x) x P(x) x P(x) at least one true ALOT P(x) x P(x) x P(x) 2. Conclusions from Quantified Statements Question 1: If we know that P(x) is true for at least one x, denoted ALOT P(x), what can we say about the statement x P(x)? By definition we can conclude x P(x) T. What about x P(x)? From the contrapositive we can conclude x P(x) T. However, if we know that ALOT P(x), what can we say about the statement x P(x)? Here we cannot make any conclusion as to whether x P(x) is true or false. This is because the statement ALOT P(x) allows for x P(x) x P(x) which violates x P(x). (I.e. consider where P(x) = T for exactly one x.) More formally we only know that ALOT P(x) x P(x) ( x P(x) x P(x)), but it is not true that ALOT P(x) x P(x) nor that ALOT P(x) x P(x) x P(x). Frequently in predicate calculus proofs we will derive statements such as ALOT P(x) or AT P(x) and then have to try and make a conclusion regarding statements such as x P(x). In some cases the derived statement, ALOT/AT, etc., will allow us to make a firm conclusion regarding the quantified statement. (As in the above where ALOT P(x) x P(x)). But in other cases the derived statement will not allow us to make a firm

2 conclusion regarding the quantified statement. (As in the above where ALOT P(x) allowed no conclusion regarding x P(x)). Example 1: Consider the domain of real numbers and the predicate P(x): x > 5. We know that for x = 6, P(x) is true, so we have shown ALOT P(x). Having found ALOT P(x) we can assert x P(x). But having found ALOT P(x), we cannot assert x P(x). That is regarding the quantified statement, x P(x), we can draw no conclusion from our proof of ALOT P(x). We say that ALOT P(x) is insufficient for x P(x) or ALOT P(x) is inconclusive with respect to x P(x) If we did want to draw a conclusion regarding x P(x), we should show ALOF P(x). We can get ALOF P(x) by showing x = 4, P(x) is false. Then we can conclude x P(x) is false. To reiterate, in this example proving ALOT P(x) gave us no conclusion for the truth value of x P(x). But proving ALOF P(x) allow us to conclude that x P(x) F. The complete rules for drawing conclusions about quantified statements are summarized below. Note the cases which are inconclusive. Table: Conclusions from Quantified Statements AF P(x) x P(x) = F x P(x) = T // definition x P(x) = F // contrapositive x P(x) = T AT P(x) x P(x) = T // definition x P(x) = F x P(x) = T x P(x) = F // contrapositive ALOT P(x) x P(x) inconclusive //ALOT may have exactly one true x P(x) = F //contrapositive x P(x) = T //definition x P(x) inconclusive //ALOT maybe AT ALOF P(x) x P(x) = F x P(x) inconclusive x P(x) inconclusive x P(x) = T //contrapositive //ALOF P(x) may have exactly one false //ALOF P(x) may have AF P(x) //definition

3 3. Predicate Calculus Implication Rules for Question 2: Consider x,y in a single domain with predicates P(x) and Q(y). Let us form the conjunction P(x) Q(y). How can we quantify, P(x) Q(y), assuming that we add a quantifier {AF,AT,ALOF,ALOT} in front of each P(x) and Q(y)? For example, if we are given A.F. for P(x) and A.T. for Q(y), how can we quantify P(x) Q(y)? That is which one of the following are valid conclusions regarding P(x) Q(y) given AF P(x) and AT Q(x)? AF (P(x) Q(y)) AT (P(x) Q(y)) ALOF (P(x) Q(y)) ALOT (P(x) Q(y)) None of the above In the case of AF P(x) AT Q(y), we can conclude AF (P(x) Q(y)). Why? Because P(x) is always F and ing it with Q(y)--which is always true--still always gives us F (recall: F q F). Hence, we conclude that: AF P(x) AT Q(y) AF (P(x) Q(y)) Further, because is symmetric, (p q q p), we also conclude AT P(x) AF Q(y) AF (P(x) Q(y)) If we carefully consider all possible combinations of quantifiers for a P(x) and Q(y), we can derive the following table. In this table, as a short-hand notation, we only list the quantifiers at not the predicates; that is the short-hand: AT AF AF means: AT P(x) AF Q(y) AF (P(x) Q(y)) Table: Predicate Calculus Implication Rules for 1. AF q where q {AF,AT,ALOF,ALOT} AF //7 cases 2. AT ALOT ALOT //2 cases 3. AT AT AT //1 case 4. ALOF q where q {AT,ALOF,ALOT} ALOF //5 cases 5. ALOT ALOT inconclusive //1 case

4 Due to symmetry of, all the above rules are true if we swap the left and right operand of the. There are 16 combinations we can make of pairing AF/AT/etc. Checking the above and accounting for symmetry, we see these rules cover all possible cases Proofs of Predicate Calculus Implication Rules for Proof of Rule 1: The AF says for all x, P(x) is false. Earlier we showed that AF AT AF because the F always dominates (recall: F q F) in all possible combinations of P(x) and Q(y). Similarly if we keep AF P(x) but choose for Q(y) either ALOF or ALOT, then we still get AF for the complete conjunction due to domination of F. Proof of Rule 2: AT ALOT guarantees P(x) is true for all x and Q(y) is true for at least one y. This means there is at least one combination of x y where P(x) Q(y) is true. So we conclude: AT ALOT ALOT Proof of Rule 3: AT AT says P(x) is always true and Q(x) is always true therefore the conjunction is always true, i.e. AT AT AT Proof of Rule 4(a): AT ALOF guarantees P(x) is true for all x and Q(y) is false for at least one y. Which of AT,AF, ALOF and ALOT are guaranteed by this proposition? We exhaustively examine all options: ALOF: ALOF is guaranteed because AT ALOF guarantees at least one pair, P(x)=T and Q(y)=F, and T F is F. AT: AT is contradicted because we have ALOF above. AF: AF is not guaranteed because AT ALOF allows for a pair P(x)=T and Q(y)=T and T T is T. ALOT: ALOT is not guaranteed because AT ALOF allows for a pair P(x)=T and for Q(y)=F for all y and T F is F. Therefore we conclude: AT ALOF ALOF Proof of Rule 4(b): ALOT ALOF says P(x) is true for at least one x and Q(y) is false for at least one y. Which of AT,AF, ALOF and ALOT are guaranteed by this proposition? We exhaustively examine all options: ALOF: ALOF is guaranteed because ALOT ALOF guarantees that Q(y)=F for at least one y. P(x) F is F for all x. AT: AT is contradicted because we have ALOF above.

5 AF: AF is not guaranteed because ALOT ALOF allows for a pair P(x)=T and Q(y)=T and T T is T. ALOT: ALOT is not guaranteed because ALOT ALOF allows that Q(y)=F for all y and P(x) F is F for any x. So we conclude: ALOT ALOF ALOF Proof of Rule 4(c): ALOF ALOF says P(x) is false at least once and Q(y) is false at least once. ALOF: ALOF is guaranteed because ALOF ALOF guarantees that Q(y)=F for at least one y. P(x) F is F regardless of the true value of P(x). AT: AT is contradicted because we have ALOF above. AF: AF is not guaranteed because ALOF ALOF allows for some P(x) to be T and some Q(y) to be T. T T is T which violates AF. ALOT: ALOT is not contradicted but is not guaranteed because ALOF ALOF allows for P(x) to be always F and Q(y) to be always F which yields a conjunction which always false, violating ALOT. So we conclude: ALOF ALOF ALOF Proof of Rule 5: ALOT ALOT says P(x) is true at least once and Q(y) is true at least once. AT: AT is not guaranteed because ALOT ALOT allows for a P(x) to F for some x this makes the conjunction false regardless of the truth value of Q(y). This violates AT. AF: AF is not guaranteed because ALOT ALOT allows P(x) to be T for all x and Q(y) to be T for all y which would cause the conjunction to be AT, which violates AF. ALOT: ALOT is not guaranteed because ALOT ALOT allows the case where P(x) is T when Q(y) is F and P(x) is F when Q(y) is T. Both cases yield a false conjunction; hence ALOT ALOT allows a conjunction which is AF. ALOF: ALOF is not guaranteed because ALOT ALOT allows P(x) to T for all x and Q(y)=T for all y which could create a conjunction of AT, violating ALOF. Hence ALOT ALOT does not individually guarantee any of AT,AF,ALOT or ALOF. We denote this by writing: ALOT ALOT is inconclusive or just: ALOT ALOT?

6 [Strictly speaking the above is not a well-formed logic statement. More formally, the above should be: ALOT ALOT err.. I don t know]. 4. Predicate Calculus Implication Rules for Table: Predicate Calculus Implication Rules for 1. AT q where q {AF,AT,ALOF,ALOT} AT //7 cases 2. AF ALOF ALOF //2 cases 3. AF AF AF //1 case 4. ALOT q where q {AF,ALOF,ALOT} ALOT //5 cases 5. ALOF ALOF inconclusive //1 case Due to symmetry of, all the above rules are true if we swap the left and right operand of the. There are 16 combinations we can make of pairing AF/AT/etc. Checking the above and accounting for symmetry, we see these rules cover all possible cases Proof: Predicate Calculus Implication Rules for We could follow a similar pattern of proofs for as we did for. However, from the duality principle (see textbook), we can directly derive the table from the earlier table by swapping for and swapping all T and F s.

7 Definitions: AF,AT,ALOF,ALOT Definitions: AF,AT,ALOF,ALOT Predicate Calculus Axioms (Cheatsheet) Syntax Definition Contrapositive all false AF P(x) x P(x) x P(x) all true AT P(x) x P(x) x P(x) at least one false ALOF P(x) x P(x) x P(x) at least one true ALOT P(x) x P(x) x P(x) Conclusions from Quantified Statements AF P(x) x P(x) = F x P(x) = T // definition x P(x) = F x P(x) = T AT P(x) x P(x) = T // definition x P(x) = F x P(x) = T x P(x) = F ALOT P(x) x P(x) inconclusive //ALOT allows for exactly one true x P(x) = F //opposite x P(x) = T //definition x P(x) inconclusive //ALOT allows for AT ALOF P(x) x P(x) = F x P(x) inconclusive x P(x) inconclusive x P(x) = T //opposite //ALOF P(x) allows for exactly one false //ALOF P(x) allows for AF P(x) //definition Implication Rules for 1. AF q where q {AF,AT,ALOF,ALOT} AF 2. AT ALOT ALOT 3. AT AT AT 4. ALOF q where q {AT,ALOF,ALOT} ALOF 5. ALOT ALOT inconclusive Due to symmetry of, all the above rules are true if we swap the left and right operand of the.

8 Table: Predicate Calculus Implication Rules for 1. AT q where q {AF,AT,ALOF,ALOT} AT //7 cases 2. AF ALOF ALOF //2 cases 3. AF AF AF //1 case 4. ALOT q where q {AF,ALOF,ALOT} ALOT //5 cases 5. ALOF ALOF inconclusive //1 case Due to symmetry of, all the above rules are true if we swap the left and right operand of the.