EE1J2 - Slide 1 EE1J2 – Discrete Maths Lecture 11 Introduction to Predicate Logic Limitations of Propositional Logic Predicates, quantifiers and propositions NL interpretation of statements in Predicate Logic Formalisation in Predicate Logic of statements in NL Some equivalences

EE1J2 - Slide 2 Limitations of propositional logic Consider the following statements All French people speak French Some French people speak French There are people in France who don’t speak French Although there are clear relations between these statements, these are not exposed by propositional logic

EE1J2 - Slide 3 Limitations of propositional logic (continued) Consider also: ‘person x’ speaks French This is not an elementary proposition because its truth or falsehood depends on x

EE1J2 - Slide 4 Examples from arithmetic There are many examples of this type in arithmetic: 7  3 = 21 7  3 = 23 are both elementary propositions, but: 7  x = 21 is not, because its truth or falsehood depends on the value of x

EE1J2 - Slide 5 Predicates In each of the examples, truth or falsehood depends on whether or not a particular property is satisfied In formal logic we use the term predicate to indicate a property which a variable may or may not have

EE1J2 - Slide 6 Example Predicates From the first examples we have the predicates: P(x) = ‘x is French’ Q(x) = ‘x speaks French’ In the example from arithmetic R(x) = ‘7x = 21’ S(x) = ‘sin(x)  1’ These are examples of one-place, or unary predicates – I.e. they have just one argument

EE1J2 - Slide 7 N -place Predicates There are also two-place, or binary predicates, e.g: P(x,y) = ‘3x + 5y = 25’ Q(x,y) = ‘sin(x) = sin(y)’ and 3-place predicates: R(x,y,z) = ‘6x + 3y 2 + z = 21’ S(x,y,z) = ‘x 2 + y 2  z 2 and, in general, N-place predicates

EE1J2 - Slide 8 Instantiation Once we choose values for its variables, a predicate becomes a proposition For example: R(x) = ‘7x = 21’ (from slide 6). So R(3) is the proposition ‘7  3=21’, which is true, but R(4) is the proposition ‘7  4=21’ which is false P(x,y) = ‘3x + 5y = 25’ (from slide 7). So P(6,7) is the proposition ‘3   7 = 25’, which is false

EE1J2 - Slide 9 Quantifiers Statements involving predicates can also be made into propositions using quantifiers:  x ‘for all x’, ‘for every x’  x ‘there exists an x’, ‘for some x’ For example, if P(x) = ‘sin(x)  1’, then  x:P(x) is a proposition, which is true

EE1J2 - Slide 10 Example 1 As a second example, suppose P(x,y) = ‘3x + 5y = 25’ then  x  y:P(x,y) is a proposition: “There exists x and there exists y such that 3x+5y=25” The proposition is true: e.g. take x=5 and y=2

EE1J2 - Slide 11 Example 2 Now consider:  x  y: P(x,y) In ‘natural language’the proposition is: “For every integer x there exists an integer y such that 3x+5y=25” Assuming that the values of x and y are integers, the proposition is false (why?)

EE1J2 - Slide 12 Example 3 Now consider the 3-place predicate P(x,y,z) = ‘x 2 + 5y + z 3 = 27’ and consider the statement:  x  y : P(x,y,z) This is not a proposition (why?)

EE1J2 - Slide 13 NL interpretation of statements in Predicate Logic Consider the following predicates: P(x) = ‘x is French’, Q(x) = ‘x was born in Paris’ R(x) = ‘x speaks French’, S(x,y) = ‘x is related to y’ Express the following as statements in ‘NL’  x : P(x)  Q(x) “For every x, if x is French then x was born in Paris’ “Everyone who is French was born in Paris” (Better!) “All French people were born in Paris”

EE1J2 - Slide 14 NL interpretation (continued) P(x) = ‘x is French’, Q(x) = ‘x was born in Paris’ R(x) = ‘x speaks French’, S(x,y) = ‘x is related to y’  x  y : P(x)  P(y)  S(x,y) “For every person x there is a person y such that x is French and y is French and x and y are not related” “For every French person there is another French person who they are not related to” “No French person is related to every other French Person”

EE1J2 - Slide 15 NL Interpretation P(x) = ‘x is French’, Q(x) = ‘x was born in Paris’ R(x) = ‘x speaks French’, S(x,y) = ‘x is related to y’  x :  Q(x)   R(x)  y : P(y)  (  x(Q(x)  S(x,y))  R(x))  x  y : (P(x)  P(y))  S(x,y)

EE1J2 - Slide 16 Formalisation of NL P(x) = ‘x is French’, Q(x) = ‘x was born in Paris’ R(x) = ‘x speaks French’, S(x,y) = ‘x is related to y’ “Everyone who was born in Paris speaks French”  x : Q(x)  R(x) “There is at least one French person whose relatives all speak French”  x : P(x)  (  y (S(x,y)  R(y))

EE1J2 - Slide 17 Formalisation of NL (cont.) P(x) = ‘x is French’, Q(x) = ‘x was born in Paris’ R(x) = ‘x speaks French’, S(x,y) = ‘x is related to y’ “There is a French person all of whose relatives are also French” “All French people who were not born in Paris cannot speak French” “There is at least one French person who was neither born in Paris nor speaks French”

EE1J2 - Slide 18 Proofs involving  To show that the statement  xP(x) is true, we have to show that P(a) is true for every possible a Conversely, if P(a) is true for every possible a then  xP(x) is true

EE1J2 - Slide 19 Proofs involving  To show that the statement  xP(x) is true, we have to show that P(a) is true for at least one a Conversely, if P(a) is true for some a then  xP(x) is true

EE1J2 - Slide 20 Some equivalences Let P be a predicate Suppose  (  xP(x)) Then it is not true that P(x) holds for every x So, there must be at least one x for which P(x) is false So  x(  P(x)) is true Hence  (  xP(x))   x(  P(x))

EE1J2 - Slide 21 Equivalences (continued) Suppose  (  xP(x)) Then it is not true that P(x) holds for at least one x So, P(x) is false for every x So  x(  P(x)) is true Hence  (  xP(x))   x(  P(x))

EE1J2 - Slide 22 More equivalences  x(P(x)  Q(x))   xP(x)  xQ(x)  x(P(x)  Q(x))   xP(x)   xQ(x) If  xP(x)   xQ(x) then  x(P(x)  Q(x)) If  x(P(x)  Q(x)) then  xP(x)   xQ(x)

EE1J2 - Slide 23 Summary Introduction to predicate logic Introduction to N-place predicates Turning predicates into propositions using instantiation or quantifiers NL interpretation of statements in predicate logic (and vice versa) Some equivalences