Question

Use structural induction to prove that$\mathbf{(}\mathbf{w}\mathbf{1}\mathbf{w}\mathbf{2}{\mathbf{)}}^{\mathbf{R}}\mathbf{=}{\mathbf{w}}_{\mathbf{2}}^{\mathbf{R}}{\mathbf{w}}_{\mathbf{1}}^{\mathbf{R}}$

Short answer

It is proved that$(\mathrm{w}1\mathrm{w}2{)}^{\mathrm{R}}={\mathrm{w}}_2^{\mathrm{R}}{\mathrm{w}}_1^{\mathrm{R}}$

Step-by-step solution

Introduction

Structural induction is defined as the method of proof that is used in mathematical logic computer science, graph theory, and some other mathematical fields.

Step 2: Prove (w1w2)R=w2Rw1R

Individuals shall prove ${\left(w_1{W}_2\right)}^R=w_2^R{W}_1^R$using structural induction and individuals first define the reversal function R,

$c^R=c\text{and}(wx{)}^R=x{w}^R$

Basic step:

We have $w_2=c$ then,

$\begin{array}{r}{\left({\mathrm{w}}_1\mathrm{c}\right)}^{\mathrm{R}}={\mathrm{w}}_1^{\mathrm{R}}{\mathrm{c}}^{\mathrm{R}}\left[\text{Use defination}(\mathrm{wx}{)}^{\mathrm{R}}={\mathrm{xw}}^{\mathrm{R}}\right]\\ ={\mathrm{w}}_1^{\mathrm{R}}\mathrm{c}\\ ={\mathrm{cw}}_1^{\mathrm{R}}\\ ={\mathrm{c}}^{\mathrm{R}}{\mathrm{w}}_1^{\mathrm{R}}\left[\text{Use defination}{\mathrm{c}}^{\mathrm{R}}=\mathrm{c}\right]\end{array}$

Individuals assume that ${\left(W_1{W}_2\right)}^R=W_2^R{W}_1^R$

Recursive step:

Now we have to prove,

$\begin{array}{r}{\left(w_1{w}_{2}v\right)}^R=v{\left(w_{2}v\right)}^R\left[\text{Use}\left(w{x}^R\right)=x{w}^R\right]\\ =v{w}_2^R{w}_1^R\left[\text{Use our assumption}\right]\\ =\left(v{w}_1^R\right)w_1^R\left[\text{Use associative property}\right]\\ ={\left(w_{2}v\right)}^R{w}_1^R\left[\text{Use}\left(w{x}^R\right)=x{W}^R\right]\end{array}$

Hence, it is proved${\left.({\mathrm{w}}_1{\mathrm{w}}_2\mathrm{v}\right)}^{\mathrm{R}}={\left({\mathrm{w}}_2\mathrm{v}\right)}^{\mathrm{R}}{\mathrm{w}}_1^{\mathrm{R}}$