$a < b$ and $c<d$ imply $a+c < b+d$
$a < b$ and $c<d$ imply $a+c < b+d$ when $a,b,c,d$ are arbitrary
nonnegative integers.
I know that (assuming we include zero)
$$begin{align*}
a<b Leftrightarrow (exists xin mathbb N)a + S(x) = b\
c<d Leftrightarrow (exists yin mathbb N)c + S(y) = d
end{align*}$$
And that's what I have done by using associative and conmutative properties:
$$(a+c)+(S(x)+S(y))\=a+(c+S(x))+S(y)\=a + (S(x)+c)+S(y)\=(a+S(x)) + (c+S(y))\= b + d$$
Also, we have that $S(x) + S(y) = S(S(x+y))$ by using addition definition recursively.
$x land y in mathbb{N} implies S(x) land S(y) in mathbb{N}^*$
where $mathbb N^* =mathbb Nsetminus{0}$
That means we could start with $(a + c) + 1 = b + d$ and could work with any $k in mathbb N^*$ that satisfies the expression for arbitrary $a, c, b, d in mathbb N$, thus $a+c < b+d$.
Is that proof valid in the context of naturals and Peano axioms?
proof-verification inequality proof-writing peano-axioms natural-numbers
asked Dec 7 at 15:31
adriana634
416
|
show 2 more comments
$a < b$ and $c<d$ imply $a+c < b+d$ when $a,b,c,d$ are arbitrary
nonnegative integers.
I know that (assuming we include zero)
$$begin{align*}
a<b Leftrightarrow (exists xin mathbb N)a + S(x) = b\
c<d Leftrightarrow (exists yin mathbb N)c + S(y) = d
end{align*}$$
And that's what I have done by using associative and conmutative properties:
$$(a+c)+(S(x)+S(y))\=a+(c+S(x))+S(y)\=a + (S(x)+c)+S(y)\=(a+S(x)) + (c+S(y))\= b + d$$
Also, we have that $S(x) + S(y) = S(S(x+y))$ by using addition definition recursively.
$x land y in mathbb{N} implies S(x) land S(y) in mathbb{N}^*$
where $mathbb N^* =mathbb Nsetminus{0}$
That means we could start with $(a + c) + 1 = b + d$ and could work with any $k in mathbb N^*$ that satisfies the expression for arbitrary $a, c, b, d in mathbb N$, thus $a+c < b+d$.
Is that proof valid in the context of naturals and Peano axioms?
proof-verification inequality proof-writing peano-axioms natural-numbers
asked Dec 7 at 15:31
adriana634
416
2
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
2
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34
|
show 2 more comments
$a < b$ and $c<d$ imply $a+c < b+d$ when $a,b,c,d$ are arbitrary
nonnegative integers.
I know that (assuming we include zero)
$$begin{align*}
a<b Leftrightarrow (exists xin mathbb N)a + S(x) = b\
c<d Leftrightarrow (exists yin mathbb N)c + S(y) = d
end{align*}$$
And that's what I have done by using associative and conmutative properties:
$$(a+c)+(S(x)+S(y))\=a+(c+S(x))+S(y)\=a + (S(x)+c)+S(y)\=(a+S(x)) + (c+S(y))\= b + d$$
Also, we have that $S(x) + S(y) = S(S(x+y))$ by using addition definition recursively.
$x land y in mathbb{N} implies S(x) land S(y) in mathbb{N}^*$
where $mathbb N^* =mathbb Nsetminus{0}$
That means we could start with $(a + c) + 1 = b + d$ and could work with any $k in mathbb N^*$ that satisfies the expression for arbitrary $a, c, b, d in mathbb N$, thus $a+c < b+d$.
Is that proof valid in the context of naturals and Peano axioms?
proof-verification inequality proof-writing peano-axioms natural-numbers
asked Dec 7 at 15:31
adriana634
416
$a < b$ and $c<d$ imply $a+c < b+d$ when $a,b,c,d$ are arbitrary
nonnegative integers.
I know that (assuming we include zero)
$$begin{align*}
a<b Leftrightarrow (exists xin mathbb N)a + S(x) = b\
c<d Leftrightarrow (exists yin mathbb N)c + S(y) = d
end{align*}$$
And that's what I have done by using associative and conmutative properties:
$$(a+c)+(S(x)+S(y))\=a+(c+S(x))+S(y)\=a + (S(x)+c)+S(y)\=(a+S(x)) + (c+S(y))\= b + d$$
Also, we have that $S(x) + S(y) = S(S(x+y))$ by using addition definition recursively.
$x land y in mathbb{N} implies S(x) land S(y) in mathbb{N}^*$
where $mathbb N^* =mathbb Nsetminus{0}$
That means we could start with $(a + c) + 1 = b + d$ and could work with any $k in mathbb N^*$ that satisfies the expression for arbitrary $a, c, b, d in mathbb N$, thus $a+c < b+d$.
Is that proof valid in the context of naturals and Peano axioms?
proof-verification inequality proof-writing peano-axioms natural-numbers
proof-verification inequality proof-writing peano-axioms natural-numbers
asked Dec 7 at 15:31
adriana634
416
asked Dec 7 at 15:31
adriana634
416
edited Dec 8 at 20:45
asked Dec 7 at 15:31
adriana634
416
asked Dec 7 at 15:31
adriana634
416
asked Dec 7 at 15:31
adriana634
416
416
2
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
2
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34
|
show 2 more comments
2
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
2
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34
2
2
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
2
2
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34
|
show 2 more comments
1 Answer
1
active
oldest
votes
As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$
answered Dec 9 at 20:30
Bram28
59.7k44186
add a comment |
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As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$
answered Dec 9 at 20:30
Bram28
59.7k44186
add a comment |
As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$
answered Dec 9 at 20:30
Bram28
59.7k44186
add a comment |
As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$
answered Dec 9 at 20:30
Bram28
59.7k44186
As pointed out in the comments, your proof is not quite complete.
In order to tshow that $a + c < b + d$, you need to show that there is something $z$ such that $$(b + d) + S(z) = a + c$$
all you have done is to show that
$$(b + d) + (S(x) + S(y)) = a + c$$
Fortunately, this problem is easily rectified, since you can show that $$S(x) + S(y) = S(S(x) + Y)$$
And hence you have the $z$ as need: $S(x) + y$
answered Dec 9 at 20:30
Bram28
59.7k44186
answered Dec 9 at 20:30
Bram28
59.7k44186
answered Dec 9 at 20:30
Bram28
59.7k44186
answered Dec 9 at 20:30
Bram28
59.7k44186
59.7k44186
add a comment |
add a comment |
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2
You have to add that $S(x)+S(y)=S(S(x+y))$
– Federico
Dec 7 at 15:36
2
More to the point, what you've done is valid, but your conclusion is $$(a+ c) + (S(x) + S(y)) = b + d$$while what you need to conclude is $a + c < b + d$. You have steps remaining.
– Paul Sinclair
Dec 8 at 2:26
@PaulSinclair Could I use transivity here to complete the proof? I think the step remaining is the connection between $c$ and $d$ (i.e. $a < b < c < d$).
– adriana634
Dec 8 at 15:28
That is not the problem. It appears from where you started that you are defining $u < v$ by $(exists x)u + S(x) = v$. So to match that definition and conclude that $a + c < b + d$, you need to show that there is some $w$ such that $(a + c) + S(w) = b + d$. Thus, what you need to show next is that $S(x) + S(y) = S(w)$ for some $w$. Federico has given an explicit expression for $w$, but for your purposes, all you need to show is that some $w$ works, not that explicit expression.
– Paul Sinclair
Dec 8 at 15:51
@PaulSinclair I am a bit lost in what you said. How can I show that some $w$ works? By using induction? Since we are in naturals (where $a,b,c,dinmathbb{N}$) we must use induction or is not always required? I've seen that this property is proved (where $a,b,c,dinmathbb{R}$) by only using transitivity.
– adriana634
Dec 8 at 18:34