G Problem 2 (Normal Order). The goal of this problem is to identify the unique redex, if any, that will be reduced first in normal order evaluation strategy for the given expressions. To identify a redex, you should highlight the ( λx. t ) part of the redex in yellow and the t’ part of the redex in blue. Here is an example. Example. ( λx. x x ) ( λx. x x ) ( λx. x x ) ( λx. x x ) Answer. ( λx. x x ) ( λx. x x ) ( λx. x x ) ( λx. x x ) Note that you are only asked to identify the redexes and you are not asked to do the reduction. 1. x ( λx. x x ) ( λx. x x ) x 2. ( λx. ( λx. x ) x ) ( λx. x ) x 3. ( λx. ( λx. x ) x ) λx. ( λx. x ) x 4. ( λx. x ( λx. x ) x ) λx. x ( λx. x ) x 5. x ( λx. x ( λx. x ) x ) (λx. x ( λx. x ) x ) x 6. λx. (λx. x ) (λx. ( λx. x ) x x ) 7. ( ( λx. λy. ( λy. x ) λx. x x ) ) ( ( λx. ( λy. x ) ) λx. x x ) 8. ( λx. ( λx. ( λx. x ) x ) x ) ( ( λx. x ) ( λx. x ) ( λx. x ) ) 9. x ( λx. ( λy. x ) λx. x x ) ( λx. ( λy. x ) λx. x x ) ( ( λx. ( λy. x ) ) λx. x x ) 10. ( λx. ( λy. x ) λx. x x ) ( ( λx. ( λy. x ) ) λx. x x )

answer: i realy don't get it that much can you me figer it out pleas you

step-by-step explanation:

b mark me branliest thx

step-by-step explanation:

well the answer to that is xoxoxo

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.