Some Properties of Propositional Connectors

1
Some Properties of Propositional Connectors

• Commutative properties
• S1 \/ S2 is logically equivalent to ( ltgt) S2 \/
  S1
• S1 /\ S2 ltgt S2 /\ S1
• Distributive properties
• S1 \/ (S2 /\ S3) ltgt (S1 \/ S2) /\ (S1 \/ S3)
• S1 /\ (S2 \/ S3) ltgt (S1 /\ S2) \/ (S1 /\ S3)
• Associative properties
• S1 \/ (S2 \/ S3) ltgt (S1 \/ S2) \/ S3
• S1 /\ (S2 /\ S3) ltgt (S1 /\ S2) /\ S3

2
Commutative Properties
S1
S2
S1 \/ S2
S2 /\ S1
S2 \/ S1
S1 /\ S2
T
T
T
T
T
T
T
F
T
T
F
F
T
F
T
T
F
F
F
F
F
F
F
F
3
Distributive Properties
S1
S2
S2 /\ S3
S1 \/ S2
S1\/(S2 /\ S3)
S3
(S1\/S2)/\(S1\/S3)
S1 \/ S3
T
T
T
T
T
T
T
T
F
T
T
T
F
T
T
T
F
T
T
F
T
T
T
T
T
T
T
F
F
F
T
T
T
T
T
F
T
T
T
T
T
F
F
F
T
F
F
T
F
F
F
F
T
F
F
T
F
F
F
F
F
F
F
F
F
You try the other S1/\(S2\/S3) ltgt
(S1/\S2)\/(S1/\S3)
4
Associative Properties
S1
S2
S2 \/ S3
S1 \/ S2
S1\/(S2 \/ S3)
S3
(S1\/S2)\/S3
T
T
T
T
T
T
T
F
T
T
T
T
T
T
F
T
T
T
T
T
T
T
T
T
F
F
F
T
T
T
T
F
T
T
T
T
F
T
T
T
F
T
T
F
F
F
T
T
T
T
F
F
F
F
F
F
F
You try the other S1/\(S2/\S3) ltgt (S1/\S2)/\S3
5
Number of Truth Table Entries

• Note that when we have one proposition, it can
  take on the value of either T or F. So there will
  be 2 rows of entries in the truth table.
• When there are 2 propositions, there are 4 rows
  of entries of truth table covering the 4
  combinations of truth values.
• In general, for n propositions there are 2n
  number of truth table entries covering the 2n
  combinations of truth values for the n
  propositions.

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Example with 2 propositions
S1
S2
S1gtS2
S1
S1 \/ (S1gtS2)
T
T
F
T
T
T
F
F
F
F
F
T
T
T
T
F
F
T
T
T
Note that the truth table has only 22 4 rows
of entry regardless of the compound proposition
formulated with connectors because there are only
4 combinations of truth values for 2 propositions.
7
Tautology Contradiction

• Tautology A statement that is true for all
  possibilities.
• Example
• S1 \/ S1 always has a truth value of T(rue)
• Contradiction A statement that is false for all
  possibilities.
• Example
• S1 /\ S1 always has a truth value of F(alse)

S1
S1
S1 v S1
F
T
T
F
T
T
All True
S1
S1
S1 ?S1
F
F
T
F
F
T
All False
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Why Care About Tautology ?

• It gives us a way to show equivalence (as you
  have already seen)
• It gives us a way to reason

9
Consider Several More Rules

• DeMorgans Law
• (S1 \/ S2) lt gt S1 /\ S2
• (S1 /\ S2) lt gt S1 \/ S2
• (S1) lt gt S1
• (P/\Q) \/ (P/\Q) lt gt Q
• (P\/Q) /\ (P\/Q) lt gt Q
• P \/ False lt gt P
• P /\ True lt gt P

10
Show DeMorgans Law is Valid
a
b
S1
S2
S1
S2
(S1 \/ S2)
S1 /\ S2
a lt gt b
T
T
F
F
F
F
T
T
F
F
T
F
F
T
T
F
T
F
F
F
T
F
F
T
T
T
T
T
Note that (S1\/S2) ltgt (S1 /\ S2) is all T,
a TAUTOLOGY. Thus DeMorgans Law is true or
valid You try the other DeMorgans Law (S1/\S2)
lt gt S1 \/ S2
11
Why Reasoning with Modus Ponens is Valid ?

• If P, then Q
• P
• Then Q

Because it is a Tautology !
P
Q
P gt Q
( (P gt Q) /\ P) gt Q
(P gt Q)/\ P
T
T
T
T
T
T
F
F
T
F
T
T
F
F
T
T
F
F
F
T
12
How About Modus Tollens ?

• If P, then Q
• Q
• Then P

Because it is a Tautology !
P
Q
P gt Q
( (P gt Q) /\ Q) gt P
(P gt Q)/\ Q
P
Q
T
T
T
F
F
T
F
T
F
F
T
F
T
F
T
T
F
F
T
T
F
T
F
F
T
T
T
T
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Try Disjunctive Syllogism

• P or Q
• Q
• Then P

It also is a Tautology !
P
Q
Q
(P\/Q) /\ Q
(P \/ Q)
((P\/Q) /\ Q) gt P
T
T
F
T
T
F
T
F
T
T
T
T
T
F
F
T
T
F
F
F
T
F
F
T
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Two Replacement Rules(extending a proposition
preserving its truth value)

• 1st If a compound proposition, P, is a tautology
  and if all occurrences of some variable in P, say
  r, are replaced by the same expression, say z,
  then the resulting compound proposition, P, is
  also a tautology.
• Let tautology, P, be a /\ b gt a\/ b
• then replacing all occurrences of b by b gt
  c will form a new proposition, P, which is also
  a tautology. Thus the new P, a/\(bgtc) gt a
  \/ (bgtc), is also a tautology

Show this tautology in class
Try showing this example with a truth table
15
Two Replacement Rules (cont.)(extending a
proposition preserving its truth value)

• 2nd If a compound proposition, P, contains a
  proposition Q, and if Q is replaced by a
  logically equivalent proposition Q, then the
  resulting compound proposition, P, is logically
  equivalent to P.
• Let proposition, P, be a \/ (bgtc)
• then replacing bgtc in P by cgt b will
  create P which is logically equivalent to P.

Try showing this example with a truth table
16
Proof by Using Replacement Rule(s)

• The earlier example of no work, no pay is the
  same as work if paid, where P work and Q
  pay.
• the truth table was identical for (PgtQ) and
  (QgtP)
• Is there another way to show this logical
  equality ? Try using replacement rules
• QgtP original proposition of work
  if paid
• Q \/ P replacement rule 2 (replacing
  whole expression with logical equivalent)
• P \/ Q replacement rule 2 (replacing
  whole expression with logical equivalent)
• (P) \/ Q replacement rule 2 (replacing P
  with logical equivalent)
• PgtQ replacement rule 2 (replacing whole
  expression with logical equivalent)

work if paid
no work, no pay
Using Replacement Rule 2, we have started with
one proposition and ended with a logically
equivalent proposition
17
Some Different Proof Techniques (may also be
used with replacement rule)

• Modus Ponens (Syllogism)
• P gt Q , P, then Q ,
  because ((PgtQ ) /\ P) gt Q is a tautology
• Modus Tollens
• PgtQ, Q, then P
  because ((PQ) ? Q) gt P is a tautology
• Disjunctive syllogism
• P \/ Q, P, then Q ,
  because ( (P\/Q) /\ P ) gt Q is a tautology
• Contradiction
• P gtQ show P/\QgtFalse , because
  (P/\Q gt False) ltgt (P gtQ)
• Contrapositive
• P gtQ show Q gt P, because
  PgtQ ltgt QgtP
• By Cases
• (P1 \/ P2) gt Q show (P1gtQ) /\ (P2 gtQ),
  because they are equal
• Transitive rule
• P -gtQ, Q-gtR, then P-gtR because ((
  P-gtQ) ? (Q-gtR)) -gt (P-gtR) is a tautology

Show these
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Proof by Contradiction

• We often use one case of contradiction to prove
  something. So if dont believe P? Q, then we just
  find one situation where P is true AND Q is also
  true.
• P it rained
• Q I wear a raincoat
• Then to show that if it rains, I would be
  wearing a rain coat is false, all we got to find
  is one case where it rained (P) and I did not
  wear raincoat (Q).
• The reverse is that if I can never find such as a
  situation (P AND Q) then P?Q must be true.
• I can never find (P AND Q ) is the same as
  saying (P AND Q) ? False.
• Therefore P ?Q is the same as (P AND Q) ? False.

Note this twist in words
This line of thought using proof by contradiction
can be shown with truth table ---- a little
easier to see the truth table.
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Showing Contradiction Technique
P ? Q
(P AND Q) ? F
(P ?Q) (P AND Q)?F
P AND Q
P
Q
T
T
T
F
T
T
T
F
T
F
F
T
T
T
F
T
T
F
T
F
F
T
F
T
Tautology
Equal
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Can you show proof of transitivity property
?If P ?Q and Q ? Z then P ? Z
(P ? Q) AND (Q ?Z) ? (P ?Z)
Q ? Z
P ? Z
P ? Q
P
Q
Z
T
T
T
T
T
T
T
T
T
T
F
T
F
F
T
T
F
T
F
T
T
T
T
F
F
F
T
F
T
F
T
T
T
T
T
T
F
T
F
T
F
T
T
F
F
T
T
T
T
T
F
F
F
T
T
T