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If a theory has a model, does a theory have to be consistent?
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This may be a too obvious question, but then in completeness theorem, direction is only in one direction: if a theory is consistent, then it has a model. Can we make it stronger and say that a theory has a model if and only if a theory is consistent?
logic model-theory
asked Jul 23 at 1:06
Brimos
513
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up vote
3
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This may be a too obvious question, but then in completeness theorem, direction is only in one direction: if a theory is consistent, then it has a model. Can we make it stronger and say that a theory has a model if and only if a theory is consistent?
logic model-theory
asked Jul 23 at 1:06
Brimos
513
"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This may be a too obvious question, but then in completeness theorem, direction is only in one direction: if a theory is consistent, then it has a model. Can we make it stronger and say that a theory has a model if and only if a theory is consistent?
logic model-theory
asked Jul 23 at 1:06
Brimos
513
This may be a too obvious question, but then in completeness theorem, direction is only in one direction: if a theory is consistent, then it has a model. Can we make it stronger and say that a theory has a model if and only if a theory is consistent?
logic model-theory
asked Jul 23 at 1:06
Brimos
513
asked Jul 23 at 1:06
Brimos
513
asked Jul 23 at 1:06
Brimos
513
asked Jul 23 at 1:06
Brimos
513
513
"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55
add a comment |Â
"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55
"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55
add a comment |Â
1 Answer
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Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $varphi$ then $varphi$ is true in every model of $T$. In particular, if $T$ proves $perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
answered Jul 23 at 1:13
Noah Schweber
111k9139260
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $varphi$ then $varphi$ is true in every model of $T$. In particular, if $T$ proves $perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
answered Jul 23 at 1:13
Noah Schweber
111k9139260
add a comment |Â
up vote
8
down vote
accepted
Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $varphi$ then $varphi$ is true in every model of $T$. In particular, if $T$ proves $perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
answered Jul 23 at 1:13
Noah Schweber
111k9139260
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $varphi$ then $varphi$ is true in every model of $T$. In particular, if $T$ proves $perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
answered Jul 23 at 1:13
Noah Schweber
111k9139260
Yes; the other half you're asking about is the soundness theorem, which states that if $T$ proves $varphi$ then $varphi$ is true in every model of $T$. In particular, if $T$ proves $perp$ then $T$ has no model, and so by the contrapositive any satisfiable theory is consistent.
The proof of the soundness theorem is much simpler than that of the completeness theorem: we just show that each of the clauses of our proof system match up appropriately with the definition of satisfaction.
answered Jul 23 at 1:13
Noah Schweber
111k9139260
answered Jul 23 at 1:13
Noah Schweber
111k9139260
answered Jul 23 at 1:13
Noah Schweber
111k9139260
answered Jul 23 at 1:13
Noah Schweber
111k9139260
111k9139260
add a comment |Â
add a comment |Â
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"A consistent theory has a model" is also known as Henkin's Theorem, or The Henkin Theorem.
– DanielWainfleet
Jul 23 at 2:08
@DanielWainfleet I've never heard it called that - just "the completeness theorem" (of which it's an immediate corollary). Is it referred to separately as such?
– Noah Schweber
Jul 23 at 2:50
@NoahSchweber. I just recall that name from the hand-written lecture notes from a course in Model Theory, decades ago.
– DanielWainfleet
Jul 23 at 2:55