Equality of ordered n-tuples

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In a Mathematical Introduction to Logic, Enderton says it is easy to see (by using induction on $n$ and the basic property of ordered pairs ($leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$ )) that (for all $n$) if $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ then $x_i=y_i$ for $1leq ileq n$. I'm having some difficulty doing this

---

Definitions

Enderton defines $leftlangle xrightrangle:= x$ so the base case $n=1$ is trivial. He also defines $leftlangle x_1,ldots,x_n+1rightrangle :=leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle$.

---

Attempt:

For the induction step, we want to show $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle $

By definition, we have

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the basic property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$

Thus
$x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$

So I've proved
$$ leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle iff x_n+1=y_n+1 land leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$$

the induction hypothesis is $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$, but I don't see how to derive the other premise: $x_n+1=y_n+1$, or make the induction step.

---

Attempt #2
For the induction hypothesis, we assume there exists $ninmathbbN$ such that:

$leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow x_i=y_i(1leq ileq n)$

In the induction step, we show that this implies:

$leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle rightarrow x_i=y_i(1leq ileq n+1)$.

Thus for the induction step, we may assume

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the defining property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$.

Thus $x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ . Thus by the induction hypothesis, we have $x_i=y_i(1leq ileq n)$. Since we showed $x_n+1=y_n+1$, we have $x_i=y_i(1leq ileq n+1)$ which is what we wanted to prove.

elementary-set-theory induction

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edited Jul 20 at 21:10

asked Jul 20 at 20:11

Evan Rosica

491212

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In a Mathematical Introduction to Logic, Enderton says it is easy to see (by using induction on $n$ and the basic property of ordered pairs ($leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$ )) that (for all $n$) if $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ then $x_i=y_i$ for $1leq ileq n$. I'm having some difficulty doing this

---

Definitions

Enderton defines $leftlangle xrightrangle:= x$ so the base case $n=1$ is trivial. He also defines $leftlangle x_1,ldots,x_n+1rightrangle :=leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle$.

---

Attempt:

For the induction step, we want to show $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle $

By definition, we have

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the basic property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$

Thus
$x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$

So I've proved
$$ leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle iff x_n+1=y_n+1 land leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$$

the induction hypothesis is $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$, but I don't see how to derive the other premise: $x_n+1=y_n+1$, or make the induction step.

---

Attempt #2
For the induction hypothesis, we assume there exists $ninmathbbN$ such that:

$leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow x_i=y_i(1leq ileq n)$

In the induction step, we show that this implies:

$leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle rightarrow x_i=y_i(1leq ileq n+1)$.

Thus for the induction step, we may assume

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the defining property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$.

Thus $x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ . Thus by the induction hypothesis, we have $x_i=y_i(1leq ileq n)$. Since we showed $x_n+1=y_n+1$, we have $x_i=y_i(1leq ileq n+1)$ which is what we wanted to prove.

elementary-set-theory induction

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edited Jul 20 at 21:10

asked Jul 20 at 20:11

Evan Rosica

491212

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

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In a Mathematical Introduction to Logic, Enderton says it is easy to see (by using induction on $n$ and the basic property of ordered pairs ($leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$ )) that (for all $n$) if $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ then $x_i=y_i$ for $1leq ileq n$. I'm having some difficulty doing this

---

Definitions

Enderton defines $leftlangle xrightrangle:= x$ so the base case $n=1$ is trivial. He also defines $leftlangle x_1,ldots,x_n+1rightrangle :=leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle$.

---

Attempt:

For the induction step, we want to show $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle $

By definition, we have

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the basic property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$

Thus
$x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$

So I've proved
$$ leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle iff x_n+1=y_n+1 land leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$$

the induction hypothesis is $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$, but I don't see how to derive the other premise: $x_n+1=y_n+1$, or make the induction step.

---

Attempt #2
For the induction hypothesis, we assume there exists $ninmathbbN$ such that:

$leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow x_i=y_i(1leq ileq n)$

In the induction step, we show that this implies:

$leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle rightarrow x_i=y_i(1leq ileq n+1)$.

Thus for the induction step, we may assume

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the defining property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$.

Thus $x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ . Thus by the induction hypothesis, we have $x_i=y_i(1leq ileq n)$. Since we showed $x_n+1=y_n+1$, we have $x_i=y_i(1leq ileq n+1)$ which is what we wanted to prove.

elementary-set-theory induction

share|cite|improve this question

edited Jul 20 at 21:10

asked Jul 20 at 20:11

Evan Rosica

491212

In a Mathematical Introduction to Logic, Enderton says it is easy to see (by using induction on $n$ and the basic property of ordered pairs ($leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$ )) that (for all $n$) if $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ then $x_i=y_i$ for $1leq ileq n$. I'm having some difficulty doing this

---

Definitions

Enderton defines $leftlangle xrightrangle:= x$ so the base case $n=1$ is trivial. He also defines $leftlangle x_1,ldots,x_n+1rightrangle :=leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle$.

---

Attempt:

For the induction step, we want to show $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle $

By definition, we have

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the basic property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$

Thus
$x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$

So I've proved
$$ leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle iff x_n+1=y_n+1 land leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$$

the induction hypothesis is $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$, but I don't see how to derive the other premise: $x_n+1=y_n+1$, or make the induction step.

---

Attempt #2
For the induction hypothesis, we assume there exists $ninmathbbN$ such that:

$leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle rightarrow x_i=y_i(1leq ileq n)$

In the induction step, we show that this implies:

$leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle rightarrow x_i=y_i(1leq ileq n+1)$.

Thus for the induction step, we may assume

$beginalign leftlangle x_1,ldots,x_n+1rightrangle &=leftlangle y_1,ldots,y_n+1rightrangle \
leftlangle leftlangle x_1,ldots,x_nrightrangle ,x_n+1rightrangle &=leftlangle leftlangle y_1,ldots,y_nrightrangle ,y_n+1rightrangle endalign$

but the defining property of ordered pairs is that

$leftlangle x,yrightrangle =leftlangle u,vrightrangle iff x=uland y=v$.

Thus $x_n+1=y_n+1$ and $leftlangle x_1,ldots,x_nrightrangle =leftlangle y_1,ldots,y_nrightrangle$ . Thus by the induction hypothesis, we have $x_i=y_i(1leq ileq n)$. Since we showed $x_n+1=y_n+1$, we have $x_i=y_i(1leq ileq n+1)$ which is what we wanted to prove.

elementary-set-theory induction

share|cite|improve this question

edited Jul 20 at 21:10

asked Jul 20 at 20:11

Evan Rosica

491212

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 21:10

edited Jul 20 at 21:10

edited Jul 20 at 21:10

asked Jul 20 at 20:11

Evan Rosica

491212

asked Jul 20 at 20:11

Evan Rosica

491212

asked Jul 20 at 20:11

Evan Rosica

491212

491212

add a comment | 

add a comment | 

1 Answer
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up vote
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Your goal for the "induction step" is the wrong way around. Since $left<x_1,ldotsright>=left<y_1,ldotsright>$ is a premise of the implication you want to prove, the right induction step should go as:

1. Our goal is to show that if $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$ then all of the $x_i$s and $y_i$ agree.




2. Thus, assume $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$.




3. The induction hypothesis tells us that if $left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>$ then the first $n$ values agree.




4. Now, if we can show $$left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right> to left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
  – Evan Rosica
  Jul 20 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
  – Evan Rosica
  Jul 20 at 21:26
  
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
1

active

oldest

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active

oldest

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up vote
3
down vote

Your goal for the "induction step" is the wrong way around. Since $left<x_1,ldotsright>=left<y_1,ldotsright>$ is a premise of the implication you want to prove, the right induction step should go as:

1. Our goal is to show that if $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$ then all of the $x_i$s and $y_i$ agree.




2. Thus, assume $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$.




3. The induction hypothesis tells us that if $left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>$ then the first $n$ values agree.




4. Now, if we can show $$left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right> to left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
  – Evan Rosica
  Jul 20 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
  – Evan Rosica
  Jul 20 at 21:26
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

Your goal for the "induction step" is the wrong way around. Since $left<x_1,ldotsright>=left<y_1,ldotsright>$ is a premise of the implication you want to prove, the right induction step should go as:

1. Our goal is to show that if $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$ then all of the $x_i$s and $y_i$ agree.




2. Thus, assume $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$.




3. The induction hypothesis tells us that if $left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>$ then the first $n$ values agree.




4. Now, if we can show $$left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right> to left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
  – Evan Rosica
  Jul 20 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
  – Evan Rosica
  Jul 20 at 21:26
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Your goal for the "induction step" is the wrong way around. Since $left<x_1,ldotsright>=left<y_1,ldotsright>$ is a premise of the implication you want to prove, the right induction step should go as:

1. Our goal is to show that if $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$ then all of the $x_i$s and $y_i$ agree.




2. Thus, assume $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$.




3. The induction hypothesis tells us that if $left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>$ then the first $n$ values agree.




4. Now, if we can show $$left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right> to left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

Your goal for the "induction step" is the wrong way around. Since $left<x_1,ldotsright>=left<y_1,ldotsright>$ is a premise of the implication you want to prove, the right induction step should go as:

1. Our goal is to show that if $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$ then all of the $x_i$s and $y_i$ agree.




2. Thus, assume $left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right>$.




3. The induction hypothesis tells us that if $left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>$ then the first $n$ values agree.




4. Now, if we can show $$left<x_1,ldots,x_n+1right> =left<y_1,ldots,y_n+1right> to left<x_1,ldots,x_nright> =left<y_1,ldots,y_nright>,$$ then together with (2) that will give us what whe need to apply the induction hypothesis. ...

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 20:32

Henning Makholm

226k16291519

answered Jul 20 at 20:32

Henning Makholm

226k16291519

answered Jul 20 at 20:32

Henning Makholm

226k16291519

226k16291519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
  – Evan Rosica
  Jul 20 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
  – Evan Rosica
  Jul 20 at 21:26
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
  – Evan Rosica
  Jul 20 at 21:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
  – Evan Rosica
  Jul 20 at 21:26
  
  

  
  
  
  

Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
– Evan Rosica
Jul 20 at 21:16

Thanks for your answer. It helped me realize that the source of my confusion was that the induction hypothesis was itself an implication, so the logic of the induction step was $(A rightarrow B) rightarrow (C rightarrow D)$. I've added another attempt. Could you please comment? If everything is good i'll accept this answer.
– Evan Rosica
Jul 20 at 21:16

And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
– Evan Rosica
Jul 20 at 21:26

And $(A rightarrow B) rightarrow (C rightarrow D) equiv [C land (A rightarrow B)] rightarrow D$. In this case, $C$ was $leftlangle x_1,ldots,x_n+1rightrangle =leftlangle y_1,ldots,y_n+1rightrangle$ and $A rightarrow B$ was the induction hypothesis, which explains how those two facts gave us the result.
– Evan Rosica
Jul 20 at 21:26

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