ENGINEERING COLLEGE

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

Answers

Answer 1

Answer:

Answer:

Decrease to typical from utilizing lambda-decrease:

The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2

The of taking the terms is significant in lambda - math,

For the term, (λy, y×3)2, we can substitute the incentive to the capacity.

Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6

Presently the tem becomes, (λf λx f(f(fx)))6

The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.

Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.

In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.

Related Questions

What is the function of deaerator in thermal power plant?
Make a python code.Write a function named max that accepts two integer values as arguments and returns the value that is the greater of the two. For example, if 7 and 12 are passed as arguments to the function, the function should return 12. Use the function in a program that prompts the user to enter two integer values. The program should display the value that is the greater of the two. Write the program as a loop that continues to prompt for two numbers, outputs the maximum, and then goes back and prompts againHere’s an example of program useInput the first number: 10Input the second number: 5The maximum value is 10Run again? yesInput the first number: -10Input the second number: -5The maximum value is -5Run again? noFunction max():Obtain two numbers as input parameters: max(num1, num2):if num1 > num2 max_val = num1, else max_val = num2return max_valMain Program:Initialize loop control variable (continue = ‘y’)While continue == ‘y’Prompt for first numberPrompt for second numberCall function "max," sending it the values of the two numbers, capture result in an assignment statement:max_value = max (n1, n2)Display the maximum value returned by the functionprint(‘Max =’, max_val)Ask user for if she/he wants to continue (continue = input(‘Go again? y if yes’)
Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ
What’s are the preventive strategies for workplace violence
A new pipeline is installed to convey 2500 gal/min. If the pipeline can not exceed 6 ft/1,000ft of head loss what is the minimum standard diameter to convey the flowrate? Size both PVC (C=130) and steel (C=120) pipelines.

The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Answers

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

Answers

Answer:

$\eta _(max) = 0.2413 = 24.13%$

$\eta' _(max) = 0.5061 = 50.61%$

Given:

$T_(1max) = 100^(\circ) = 273 + 100 = 373 K$

operating temperature of heat engine, $T_(2) = 10^(\circ) = 273 + 10 = 283 K$

$T_(3max) = 300^(\circ) = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _(max)$ is given by:

$\eta _(max) = 1 - (T_(2))/(T_(1max))$

$\eta _(max) = 1 - (283)/(373) = 0.24$

$\eta _(max) = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_(3max)$ changes to $300^(\circ)$ , so, the new maximum efficiency, $\eta' _(max)$ is given by:

$\eta' _(max) = 1 - (T_(2))/(T_(3max))$

$\eta _(max) = 1 - (283)/(573) = 0.5061$

$\eta _(max) = 0.5061 = 50.61%$

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period.
b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits.
c. Do all of these samples meet the contract specifications? Explain your answer.
d. What trend do you observe based on the data? What could cause this trend?"

Answers

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535, standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

Find the minimum sum of products expression using Quine-McCluskey method of the function. F(A,B,C,D)= Î£ m(1,5,7,8,9,13,15)+ Î£ d(4,6,11)

Answers

Answer:

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

Explanation:

Hi please follow me also I you can and thanks for the points. Have a good day.

Determine the degrees of superheat of steam at 101.325 kPa and 170°C. a. 50°C b. 70°C c. 60°C d. 80°C

Answers

Answer:

b) 70°C

Explanation:

Given that super heat temperature at 101.325 KPa= 170°C.

We know that saturation temperature at 101.325 KPa is 100°C.

Degree of super heat =Super heat temperature - saturation temperature ( at constant pressure)

Now by putting the values

Degree of super heat=170-100

Degree of super heat=70°C.

So our option b is right.

1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/sspeed zone. A police car observed the automobile. At the instant that the two
vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². The motorist noticed the police car in his rear
view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s². (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.

Answers

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.