References. Logic (Mathematics 1BA1) Introduction to Logic. Introduction to Logic. Expressions. State and Evaluation.

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1 References Logic (Mathematics 1BA1) Rozenn Dahyot Room 128 Lloyd Institute School of Computer Science and Statistics Trinity College Dublin, IRELAND Week 1 - January 2007 A Logical Approach To Discrete Math D. Gries & F. B. Schneider, Springer Logic And Discrete Mathematics - A Computer Science Perspective W. K. Grassman & J.-P. Tremblay, Prentice Introduction to Logic M. Huggard & Mark Dukes, Trinity College textbook Boolean Algebra with Computer Applications G. E. Williams, McGraw-Hill Introduction to Logic Introduction to Logic Logic is central in many areas: Maths Philosophy Linguistic Computer Science etc. Sometimes our logical reasoning is faulty and errors can result. Consequently arised the need of identifying the laws of logic. If we had some exact language...or at least a kind of truly philosophic writing, in which the ideas were reduced to a kind of alphabet of human thought, then all that follows rationally from what is given could be found by a kind of calculus, just as arithmetical or geometrical problems are solved. Leibniz ( ) The idea of machines that reason has been around for a long time. Expressions State and Evaluation Denitions. Mathematical expressions are made up of constants, variables and operators. Parentheses are used to indicate aggregation (i.e. to bring together different expressions). 3 (x + 2) = 5 The constants are 2 and 5, the variable is x, the operators are, + and =. The parentheses bring together x and 2. 3 (x + 2) denotes the product of 3 and (x + 2). Denitions. A state is a list of variables with specic values assigned to them. Evaluation of an expression E in a state means replacing all variables in E by their values in the state and simplifying the expression. If we have E as x + y + z, then E evaluated in the state (x, 1), (y, 3) and (z, 5) is simply = 9.

2 Textual Substitution Textual Substitution. If we have an expression E containing some variable x, then we can replace every occurrence of x in E with the expression (R). This operation, called textual substitution takes precedence over all other operations, and is written as E[x := R]. Let E be the expression (x + 1)(x y) and let R = y + z. Then E[x := R] = ((x + 1)(x y))[x := y + z] = (((y + z) + 1)((y + z) y)) = (y + z + 1)(y + z y) = (y + z + 1)z Exercise: Textual substitution 1 Let the variables x, y, z be in the state (x, 1), (y, 42), (z, 5). Evaluate the following expressions in this state: (i) 2x + 3y z (v) 5x z + 1 (ii) 2 + x + 4yz (vi) 7x 2 + y yz (iii) 32 (vii) xyz (iv) x + y(z x) + z x (viii) (y 2x) z 2 Perform the following substitutions: (i) x[x := 4] (v) (x + y x + z)[x := z] (ii) (x)[x := x y z] (vi) ( )[k := π] (iii) (m x + c)[x := 1] (vii) (7 + x z + u v)[u := 4] (iv) m x + c[x := 1] (viii) (e = v c)[v := m c] Remarks on Parentheses Remarks on Parentheses The order in which parentheses are placed is important. The expression m = (y 2 y 1 )/(x 2 x 1 ) gives the slope of a line with points (x 1, y 1 ) and (x 2, y 2 ). Without the parentheses we would have m = y 2 y 1 x 2 x 1 which does not give the slope. Similarily in textual substitution, E[x := R] means replace every x in E with (R). The reason that we replace x by (R) rather than R is for rigor. This way we avoid making mistakes when working with the new expression. BODMAS Brackets Order Division Multiplication Addition Subtraction. The Reverse Polish Notation BODMAS rule Simultaneous Substitution & Iterated Substitution What happens if we want to substitute more than one variable in an expression? There are two different ways we can do this: simultaneous substitution, iterated substitution. Simultaneous Substitution Simultaneous substitution is the act of replacing all the variables at once. The simultaneous substitution E[x, y := R 1, R 2 ] means replace every x by (R 1 ) and every y by (R 2 ). Example of Simultaneous substitution: If E is x + 2y + 3z, R 1 is b 2a, R 2 is a and R 3 is 0, then E[x, y, z := R 1, R 2, R 3 ] = (x + 2y + 3z)[x, y, z := b 2a, a, 0] = ((b 2a) + 2(a) + 3(0)) = (b 2a + 2a + 0) = b

3 Iterated Substitution Iterated Substitution Iterated substitution is the act of applying substitutions to an expression one-by-one. E[x := R 1 ][y := R 2 ] means rst do E[x := R 1 ] and call it E. Then do E [y := R 2 ]. Example of Iterated substitution: If E is the expression x + 2y + 3z, R 1 is 2y, R 2 is z and R 3 is 2, then E[x := R 1 ][y := R 2 ][z := R 3 ] = (x + 2y + 3z)[x := 2y][y := z][z := 2] = ((2y) + 2y + 3z)[y := z][z := 2] = ((2(z)) + 2(z) + 3z)[z := 2] = ((2((2))) + 2((2)) + 3(2)) = ( ) = 14 Note that when doing iterated substitution, the order in which the substitutions are done is important. Example of Iterated substitution: E is the expression x + 2y + 3z, R 1 is 2y, R 2 is z and R 3 is 2. Compute E[y := R 2 ][x := R 1 ][z := R 3 ] E[y := R 2 ][x := R 1 ][z := R 3 ] = (x + 2y + 3z)[y := z][x := 2y][z := 2] = (x + 2(z) + 3z)[x := 2y][z := 2] = (x + 5z)[x := 2y][z := 2] = ((2y) + 5z)[z := 2] = (2y + 10) Simultaneous & Iterated Substitution Remark on assignement := Simultaneous (x + 2y)[x, y := y, x] = Denition: x := E evaluates expression E and stores the result in variable x. Assignment x := E is read x becomes E. To increment x in a loop (program) Iterated (x + 2y)[x := y][y := x] = x := x + 1 On the other hand, to write (x = x + 1) (equivalent to 1 = 0) is false. Summary Substitution as an Inference Rule you know about: expressions, operators, variables, states, evaluation of expression, Iterated & Simultaneous Textual Substitution, Inference Rule Substitution: If E is true, then E[v := R] must be true for any expression R. E Written as E[v := R] difference between assignment and equality, Precedence of textual substitution. if E dened as x + y = y + x is a true, the substitution rule allows us to conclude that b + 3 = 3 + b is also true (this is E[x, y := b, 3]). Precedence is the characteristic of operators that indicates when they will be evaluated when they appear in complex expressions

4 4 laws of Equality Using the Inference Rule Leibniz Four laws of equality 1 Reexivity x = x 2 Symmetry (x = y) = (y = x) 3 Transitivity X = Y, Y = Z X = Z 4 Inference rule Leibniz if X = Y is true E[z := X] = E[z := Y] is true. ( ) X = Y i.e. E[z := X] = E[z := Y] Let X = 2x + 1, Y = y 3x and E be the expression (z + 1)(z 1). The Inference Rule Leibniz gives X = Y E[z := X] = E[z := Y] From this we may derive giving yielding E[z:=X] {}}{ (2x )(2x + 1 1) = 2x + 1 = y 3x E[z := 2x + 1] = E[z := y 3x] E[z:=Y] {}}{ (y 3x + 1)(y 3x 1) 2x(2x + 2) = (y 3x + 1)(y 3x 1) Exercise: Leibniz's Inference Rule Apply Inference Rule Leibniz to the following (where substitution takes place in the variable z): (i) X = x, Y = x + 2 and E = 4z + y (ii) X = x, Y = x + 2 and E = 4x + z (iii) X = 2x + 1, Y = w and E = 2x z (iv) X = 2w + 1, Y = a + 3w 5 and E = 2z 4w 10 (v) X = x, Y = x + 2 and E = 2 Solution: Leibniz's Inference Rule Computing E[z := X] = E[z := Y] gives: (i) (4z + y)[z := x] = (4z + y)[z := x + 2] gives (4x + y) = (4x y) (ii) (4x + z)[z := x] = (4x + z)[z := x + 2] gives 5x = 5x + 2 (iii) (2x z)[z := 2x + 1] = (2x z)[z := w] gives 1 = 2x + w (iv) (2z 4w 10)[z := 2w + 1] = (2z 4w 10)[z := a + 3w 5] gives 8 = 2a + 2w 20 (v) (2)[z := x] = (2)[z := x + 2] gives 2 = 2 G. W. Leibniz ( ) Function evaluation Notice that if z is a variable in an expression E, then we can dene a function f such that f(z) is the expression E. Then the textual substitution E[z := R] may be written as f(r). Denition: f.z is the new notation for f(z) Let E be the expression 2x y + m. Then we could dene f(x) = 2x y + m = E. Performing the substitution E[x := y] is the same as f(y).

5 Example and exercises Reasoning using Leibniz The purpose of this section is to Example Let f be the function f(x) = x 2 + 3x + 4. Then f.3 = f(3) = (3) 2 + 3(3) + 4 = = 22 1 Let f(x) = 2x 1 and g(x) = x 2 1. Evaluate the following (i) f.1 (v) f.g.3 (ii) g.1 (vi) g.f.3 (iii) f.3 (vii) f.g.f.1 (iv) g.10 (viii) g.g.x introduce a method for proving the equality of two expressions, and introduce a format for such proofs. The Format: Let us suppose that two expressions E 0 and E 1 are equal. A proof that both are equal will take the following format: E 0 = The reason E 1 In the angle brackets, we explain the reason or inference rule by which we derive the fact that E 0 is equal to E 1. Reasoning using Leibniz Let 1 + 2j = 3k and 2k = p. E 0 is 3 + 4j and E 1 is 1 + 3p. Show E 0 = E 1 : 3 + 4j = 1 + 2j=3k, and multiply this by k = 2k=p, and multiply this by p Reasoning using Leibniz Using Leibniz The basic form of this reasoning/proof process when using Inference Rule Leibniz is: E[z := X] = Since X = Y E[z := Y] Let E be the expression 1 + 2z. Let X = 1 + 2j and Y = 3k. Then performing Inference Rule Leibniz gives 1 + 2(1 + 2j) = 1+2j=3k 1 + 6k Assignement Statement Assignement Statement Denitions: Execution of the assignment statement x := E evaluates expression E and stores the result in the variable x. Let S be a statement. A precondition, P, of S is an assertion about the variables of S in which the statement S may be executed. A postcondition, Q, of S is an assertion about the states in which S may terminate. We represent this situation by a Hoare triple. Let our statement be x := x + 1. If x > 0 (precondition) then the state in which the statement terminates must have x > 1. Hence one Hoare triple is: {x > 0} x := x + 1 {x > 1} This Hoare triple is said to be valid because the execution of x := x + 1 in which x > 0 terminates in a state in which x > 1. {P} S {Q}

6 Assignement Statement Exercise: Assignement Statement This Hoare triple {x = 5} x := x + 1 {x = 7} is not a valid because the execution of x := x + 1 with x = 5 does not terminate in a state in which x = 7. 1 Decide whether the following triples are valid. (i) {x > 0} x := x 2 {x < 0} (ii) {x > 0} x := x 1 {x > 1} (iii) {x even} x := x/2 {x odd} Solution: Assignement Statement Conditions on Assignment Conditions on assignment. Given an assignment x := E and a postcondition R, we may write down the precondition to complete the Hoare triple: {R[x := E]} x := E {R} 1 Solution (i) non valid (ii) non valid (iii) non valid Let x := x + 1 and let R be the postcondition {x > 6}. Then our precondition is P = R[x := x + 1] = ( x > 6 )[x := x + 1] = ( (x + 1) > 6 ) = x > 5 Hence we have the Hoare triple: {x > 5} x := x + 1 {x > 6} Exercise: Conditions on Assignment Solution: Conditions on Assignment 1 Find the suitable precondition P in order to make each of the following Hoare triples valid. (i) {P} x := 2x + 1 {x 3} (ii) {P} x := 2 x {x > 15} (iii) {P} x := x + y {x = y} (iv) {P} x := 7 {x = y} 2 The following is a function dened in the C-programming language. int newfunction(int n) { int answer; answer = n*n - 50*n - 600; return answer; } What is the precondition P such that: 1 (i) x 1 (ii) x > (iii) x = 0 (iv) y = 7 ln 15 ln 2 2 x ] ; 10[ ]60;+ [ {P} x := newfunction (x) {x > 0}