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Basic theorem proving in Mathematica?
Proving uniqueness of group identity elementWhy Can't Mathematica Resolve this Simple Quantified Expression?Four color theorem in MathematicaImplicit Function Theorem
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$begingroup$
Let's say we have the following:
p is prime
n > 1 (n is an integer)
p = nq (I.e. p is a multiple of n)
It can be proved that p = n.
I've seen that Mathematica has some basic theorem proving capabilities (see the theorem-proving tag) via functions like Reduce.
Can Mathematica prove the above claim? Pointers to external resources are welcome.
theorem-proving
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's say we have the following:
p is prime
n > 1 (n is an integer)
p = nq (I.e. p is a multiple of n)
It can be proved that p = n.
I've seen that Mathematica has some basic theorem proving capabilities (see the theorem-proving tag) via functions like Reduce.
Can Mathematica prove the above claim? Pointers to external resources are welcome.
theorem-proving
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
$endgroup$
1
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!nis an integer. I've updated the post.
$endgroup$
– dharmatech
7 hours ago
3
$begingroup$
Have a look atFindEquationalProof, though I think this may be harder than it looks.
$endgroup$
– Carl Lange
6 hours ago
add a comment |
$begingroup$
Let's say we have the following:
p is prime
n > 1 (n is an integer)
p = nq (I.e. p is a multiple of n)
It can be proved that p = n.
I've seen that Mathematica has some basic theorem proving capabilities (see the theorem-proving tag) via functions like Reduce.
Can Mathematica prove the above claim? Pointers to external resources are welcome.
theorem-proving
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
$endgroup$
Let's say we have the following:
p is prime
n > 1 (n is an integer)
p = nq (I.e. p is a multiple of n)
It can be proved that p = n.
I've seen that Mathematica has some basic theorem proving capabilities (see the theorem-proving tag) via functions like Reduce.
Can Mathematica prove the above claim? Pointers to external resources are welcome.
theorem-proving
theorem-proving
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
edited 7 hours ago
dharmatech
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
asked 8 hours ago
dharmatechdharmatech
4561 gold badge9 silver badges18 bronze badges
4561 gold badge9 silver badges18 bronze badges
1
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!nis an integer. I've updated the post.
$endgroup$
– dharmatech
7 hours ago
3
$begingroup$
Have a look atFindEquationalProof, though I think this may be harder than it looks.
$endgroup$
– Carl Lange
6 hours ago
add a comment |
1
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!nis an integer. I've updated the post.
$endgroup$
– dharmatech
7 hours ago
3
$begingroup$
Have a look atFindEquationalProof, though I think this may be harder than it looks.
$endgroup$
– Carl Lange
6 hours ago
1
1
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!
n is an integer. I've updated the post.$endgroup$
– dharmatech
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!
n is an integer. I've updated the post.$endgroup$
– dharmatech
7 hours ago
3
3
$begingroup$
Have a look at
FindEquationalProof, though I think this may be harder than it looks.$endgroup$
– Carl Lange
6 hours ago
$begingroup$
Have a look at
FindEquationalProof, though I think this may be harder than it looks.$endgroup$
– Carl Lange
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Mathematica does have such a thing, though it's unfortunately not as trivial as one would hope, as that:
FindEquationalProof cannot prove theorems involving arithmetic operators by default
As such, an example:
FindEquationalProof[a == b c, a/c == b, c == 1]
Failure["PropositionFalse",
Association["MessageTemplate" -> TemplateObject[
"The proposition could not be reduced to True."
If you read the docs under possible issues a solution to work around it.
FindEquationalProof[ForAll[x, f[4*x] == 4*f[x]], ForAll[x, f[2*x] == 2*f[x]]]
(*Same error as above*)
FindEquationalProof[ForAll[a, f[mult[4, x]] == mult[4, f[x]]], ForAll[x, f[mult[2, x]] == mult[2, f[x]]], ForAll[x, y, z, mult[x, mult[y, z]] == mult[mult[x, y], z]], mult[2, 2] == 4]
As such one would have to build in the logic of multiplying for your theorem to be found.
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
$endgroup$
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Mathematica does have such a thing, though it's unfortunately not as trivial as one would hope, as that:
FindEquationalProof cannot prove theorems involving arithmetic operators by default
As such, an example:
FindEquationalProof[a == b c, a/c == b, c == 1]
Failure["PropositionFalse",
Association["MessageTemplate" -> TemplateObject[
"The proposition could not be reduced to True."
If you read the docs under possible issues a solution to work around it.
FindEquationalProof[ForAll[x, f[4*x] == 4*f[x]], ForAll[x, f[2*x] == 2*f[x]]]
(*Same error as above*)
FindEquationalProof[ForAll[a, f[mult[4, x]] == mult[4, f[x]]], ForAll[x, f[mult[2, x]] == mult[2, f[x]]], ForAll[x, y, z, mult[x, mult[y, z]] == mult[mult[x, y], z]], mult[2, 2] == 4]
As such one would have to build in the logic of multiplying for your theorem to be found.
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
$endgroup$
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
add a comment |
$begingroup$
Mathematica does have such a thing, though it's unfortunately not as trivial as one would hope, as that:
FindEquationalProof cannot prove theorems involving arithmetic operators by default
As such, an example:
FindEquationalProof[a == b c, a/c == b, c == 1]
Failure["PropositionFalse",
Association["MessageTemplate" -> TemplateObject[
"The proposition could not be reduced to True."
If you read the docs under possible issues a solution to work around it.
FindEquationalProof[ForAll[x, f[4*x] == 4*f[x]], ForAll[x, f[2*x] == 2*f[x]]]
(*Same error as above*)
FindEquationalProof[ForAll[a, f[mult[4, x]] == mult[4, f[x]]], ForAll[x, f[mult[2, x]] == mult[2, f[x]]], ForAll[x, y, z, mult[x, mult[y, z]] == mult[mult[x, y], z]], mult[2, 2] == 4]
As such one would have to build in the logic of multiplying for your theorem to be found.
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
$endgroup$
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
add a comment |
$begingroup$
Mathematica does have such a thing, though it's unfortunately not as trivial as one would hope, as that:
FindEquationalProof cannot prove theorems involving arithmetic operators by default
As such, an example:
FindEquationalProof[a == b c, a/c == b, c == 1]
Failure["PropositionFalse",
Association["MessageTemplate" -> TemplateObject[
"The proposition could not be reduced to True."
If you read the docs under possible issues a solution to work around it.
FindEquationalProof[ForAll[x, f[4*x] == 4*f[x]], ForAll[x, f[2*x] == 2*f[x]]]
(*Same error as above*)
FindEquationalProof[ForAll[a, f[mult[4, x]] == mult[4, f[x]]], ForAll[x, f[mult[2, x]] == mult[2, f[x]]], ForAll[x, y, z, mult[x, mult[y, z]] == mult[mult[x, y], z]], mult[2, 2] == 4]
As such one would have to build in the logic of multiplying for your theorem to be found.
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
$endgroup$
Mathematica does have such a thing, though it's unfortunately not as trivial as one would hope, as that:
FindEquationalProof cannot prove theorems involving arithmetic operators by default
As such, an example:
FindEquationalProof[a == b c, a/c == b, c == 1]
Failure["PropositionFalse",
Association["MessageTemplate" -> TemplateObject[
"The proposition could not be reduced to True."
If you read the docs under possible issues a solution to work around it.
FindEquationalProof[ForAll[x, f[4*x] == 4*f[x]], ForAll[x, f[2*x] == 2*f[x]]]
(*Same error as above*)
FindEquationalProof[ForAll[a, f[mult[4, x]] == mult[4, f[x]]], ForAll[x, f[mult[2, x]] == mult[2, f[x]]], ForAll[x, y, z, mult[x, mult[y, z]] == mult[mult[x, y], z]], mult[2, 2] == 4]
As such one would have to build in the logic of multiplying for your theorem to be found.
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
answered 6 hours ago
morbomorbo
7023 silver badges9 bronze badges
7023 silver badges9 bronze badges
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
add a comment |
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
$begingroup$
Excellent answer. Thank you!
$endgroup$
– dharmatech
6 hours ago
add a comment |
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1
$begingroup$
$p=5$, $q=3$, $n=5/3$, $p$ is not equal $n$. Maybe you forgot something?
$endgroup$
– yarchik
7 hours ago
$begingroup$
@yarchik Yes you're right, thank you!
nis an integer. I've updated the post.$endgroup$
– dharmatech
7 hours ago
3
$begingroup$
Have a look at
FindEquationalProof, though I think this may be harder than it looks.$endgroup$
– Carl Lange
6 hours ago