Show that $T$ is inconsistent
$begingroup$
Let $L$ be the language ${cdot, e}$ and let $T$ be the $L$-theory whose proper axioms are:
- $(forall x)(forall y)(forall z)((x.y).z=x.(y.z)) $
$(forall x)(x.e=x ,wedge , e.x=x)$
$(forall x)(exists y)(x.y=e ,wedge , y.x=e)$
$(forall x)((x.(x.(x.x)))=e)$
$(exists x)(neg (x=e) , wedge , x.(x.x)=e )$
Show that $T$ is inconsistent
My attempt:
From $5$th axiom we know $x.x.x=e$, so by $4$th axiom we get $x.e=e$ . But according to $2$nd axiom $x$ must be equal to $e$.
So, $T$ is inconsistent because $5$th axiom ,which says $(exists x)(neg (x=e))$, doesn't hold now, since $x=e$ . But I somehow doubt that this isn't a correct way of showing T is inconsistent. How can I improve the solution?
logic first-order-logic
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
$endgroup$
add a comment |
$begingroup$
Let $L$ be the language ${cdot, e}$ and let $T$ be the $L$-theory whose proper axioms are:
- $(forall x)(forall y)(forall z)((x.y).z=x.(y.z)) $
$(forall x)(x.e=x ,wedge , e.x=x)$
$(forall x)(exists y)(x.y=e ,wedge , y.x=e)$
$(forall x)((x.(x.(x.x)))=e)$
$(exists x)(neg (x=e) , wedge , x.(x.x)=e )$
Show that $T$ is inconsistent
My attempt:
From $5$th axiom we know $x.x.x=e$, so by $4$th axiom we get $x.e=e$ . But according to $2$nd axiom $x$ must be equal to $e$.
So, $T$ is inconsistent because $5$th axiom ,which says $(exists x)(neg (x=e))$, doesn't hold now, since $x=e$ . But I somehow doubt that this isn't a correct way of showing T is inconsistent. How can I improve the solution?
logic first-order-logic
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
$endgroup$
$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
1
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00
add a comment |
$begingroup$
Let $L$ be the language ${cdot, e}$ and let $T$ be the $L$-theory whose proper axioms are:
- $(forall x)(forall y)(forall z)((x.y).z=x.(y.z)) $
$(forall x)(x.e=x ,wedge , e.x=x)$
$(forall x)(exists y)(x.y=e ,wedge , y.x=e)$
$(forall x)((x.(x.(x.x)))=e)$
$(exists x)(neg (x=e) , wedge , x.(x.x)=e )$
Show that $T$ is inconsistent
My attempt:
From $5$th axiom we know $x.x.x=e$, so by $4$th axiom we get $x.e=e$ . But according to $2$nd axiom $x$ must be equal to $e$.
So, $T$ is inconsistent because $5$th axiom ,which says $(exists x)(neg (x=e))$, doesn't hold now, since $x=e$ . But I somehow doubt that this isn't a correct way of showing T is inconsistent. How can I improve the solution?
logic first-order-logic
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
$endgroup$
Let $L$ be the language ${cdot, e}$ and let $T$ be the $L$-theory whose proper axioms are:
- $(forall x)(forall y)(forall z)((x.y).z=x.(y.z)) $
$(forall x)(x.e=x ,wedge , e.x=x)$
$(forall x)(exists y)(x.y=e ,wedge , y.x=e)$
$(forall x)((x.(x.(x.x)))=e)$
$(exists x)(neg (x=e) , wedge , x.(x.x)=e )$
Show that $T$ is inconsistent
My attempt:
From $5$th axiom we know $x.x.x=e$, so by $4$th axiom we get $x.e=e$ . But according to $2$nd axiom $x$ must be equal to $e$.
So, $T$ is inconsistent because $5$th axiom ,which says $(exists x)(neg (x=e))$, doesn't hold now, since $x=e$ . But I somehow doubt that this isn't a correct way of showing T is inconsistent. How can I improve the solution?
logic first-order-logic
logic first-order-logic
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
asked Dec 24 '18 at 18:46
Leyla AlkanLeyla Alkan
1,5751724
1,5751724
$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
1
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00
add a comment |
$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
1
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00
$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
1
1
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your approach is correct; at most it needs rewording for clarity, in case a reader forgets axiom 5's existence claim applies to the same $x$ each time you invoke it. (I actually thought you'd made a mistake at first, but that's only because correct proofs can be hard to read.) Fixing an $x$ that's an example of 5, $e=x^4=ex=x$, a contradiction. As you've noted, the $=$ signs respectively use 4, 5, 2; then 5 gives the contradiction.
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
$endgroup$
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
add a comment |
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$begingroup$
Your approach is correct; at most it needs rewording for clarity, in case a reader forgets axiom 5's existence claim applies to the same $x$ each time you invoke it. (I actually thought you'd made a mistake at first, but that's only because correct proofs can be hard to read.) Fixing an $x$ that's an example of 5, $e=x^4=ex=x$, a contradiction. As you've noted, the $=$ signs respectively use 4, 5, 2; then 5 gives the contradiction.
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
$endgroup$
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
add a comment |
$begingroup$
Your approach is correct; at most it needs rewording for clarity, in case a reader forgets axiom 5's existence claim applies to the same $x$ each time you invoke it. (I actually thought you'd made a mistake at first, but that's only because correct proofs can be hard to read.) Fixing an $x$ that's an example of 5, $e=x^4=ex=x$, a contradiction. As you've noted, the $=$ signs respectively use 4, 5, 2; then 5 gives the contradiction.
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
$endgroup$
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
add a comment |
$begingroup$
Your approach is correct; at most it needs rewording for clarity, in case a reader forgets axiom 5's existence claim applies to the same $x$ each time you invoke it. (I actually thought you'd made a mistake at first, but that's only because correct proofs can be hard to read.) Fixing an $x$ that's an example of 5, $e=x^4=ex=x$, a contradiction. As you've noted, the $=$ signs respectively use 4, 5, 2; then 5 gives the contradiction.
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
$endgroup$
Your approach is correct; at most it needs rewording for clarity, in case a reader forgets axiom 5's existence claim applies to the same $x$ each time you invoke it. (I actually thought you'd made a mistake at first, but that's only because correct proofs can be hard to read.) Fixing an $x$ that's an example of 5, $e=x^4=ex=x$, a contradiction. As you've noted, the $=$ signs respectively use 4, 5, 2; then 5 gives the contradiction.
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
answered Dec 24 '18 at 18:54
J.G.J.G.
25.9k22539
25.9k22539
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
add a comment |
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
1
1
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
$begingroup$
Thanks for clarification.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:01
add a comment |
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$begingroup$
You doubt it isn't correct, or you doubt it is correct?
$endgroup$
– J.G.
Dec 24 '18 at 18:54
1
$begingroup$
I thought the way I concluded the solution was not good and needed to be improved @J.G.
$endgroup$
– Leyla Alkan
Dec 24 '18 at 19:00