Isar proof of conjunction2019 Community Moderator ElectionWhat is a good way to define a finite multiplication table in Isar?When would you use `presume` in an Isar proof?Need proof for meta-logic conjunction elimination ruleTrying to understand fix/assume/show “Failure to refine goal”; Cmd to show proof info for schematic varsHow to submit an argument to keyword proof?Taming meta implication in Isar proofsExpressing a simple declarative proof about exponents in IsarProving properties of generated listsKnowing when an Isar-style proof is actually valid in IsabelleFunctor-like construction in Isabelle/Isar

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Isar proof of conjunction



2019 Community Moderator ElectionWhat is a good way to define a finite multiplication table in Isar?When would you use `presume` in an Isar proof?Need proof for meta-logic conjunction elimination ruleTrying to understand fix/assume/show “Failure to refine goal”; Cmd to show proof info for schematic varsHow to submit an argument to keyword proof?Taming meta implication in Isar proofsExpressing a simple declarative proof about exponents in IsarProving properties of generated listsKnowing when an Isar-style proof is actually valid in IsabelleFunctor-like construction in Isabelle/Isar










0















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question



















  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago















0















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question



















  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago













0












0








0








I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.










share|improve this question
















I'm trying to prove something using Isar; so far, I've gotten to a goal that looks like this:



(∀P Q. P ≠ Q ⟶ (∃!l. plmeets P l ∧ plmeets Q l)) ∧
(∀P l. ¬ plmeets P l ⟶ (∃!m. affine_plane_data.parallel plmeets l m ∧ plmeets P m)) ∧
(∃P Q. P ≠ Q ∧ (∃R. P ≠ R ∧ Q ≠ R ∧ ¬ affine_plane_data.collinear plmeets P Q R))


(here plmeets is a function I've defined, where plmeets P l is shorthand for "the point P lies on the line l" in an affine plane, but I don't think that's important to my question.)



This goal is a conjunction of three things. I've actually already proved lemmas that seem to me to be very close to each of these things. For instance, I have



lemma four_points_a1: "P ≠ Q ⟹ ∃! l . plmeets P l ∧ plmeets Q l"


which produces the output



theorem four_points_a1: ?P ≠ ?Q ⟹ ∃!l. plmeets ?P l ∧ plmeets ?Q l


which you can see is almost exactly the first of the three conjoined items. (I admit that my other lemmas aren't quite so exactly matched to the other two items, but I'll work on that).



I'd like to say "because of lemma four_points_a1, all we have left to prove is item2 ∧ item3" and I'm pretty sure there's a way to do this. But looking at the "Programming and Proving" book doesn't suggest anything to me. In Isabelle, rather than Isar, I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.



But I cannot see how to do this in Isar.







isabelle isar






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday







John

















asked 2 days ago









JohnJohn

246214




246214







  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago












  • 1





    " I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

    – xanonec
    2 days ago












  • Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

    – John
    2 days ago







1




1





" I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

– xanonec
2 days ago






" I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal." It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.

– xanonec
2 days ago














Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

– John
2 days ago





Thanks. This appears to be the thing I needed to get me past this particular roadblock. I've posted it as a community-wiki answer so that I can "accept" it in a few days, leaving fewer unanswered questions lying around. (If you'd rather write it as an answer, I'll accept that one instead).

– John
2 days ago












1 Answer
1






active

oldest

votes


















0














According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday











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According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday
















0














According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer




















  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday














0












0








0







According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.






share|improve this answer















According to @xanonec:




I suppose I'd apply conjI twice to split the one goal into three, and then resolve the first goal.




It is possible to do this within an Isar proof. However, it may be best to use the proof method intro instead of multiple applications of rule conjI, i.e. you can use apply(intro conjI) to split the goal into 3 subgoals. Then you can use subgoal to provide a proof for each subgoal individually. However, unless you provide your entire application, it will be difficult to say whether or not there exist better methods.




According to @John:
The syntax for this process that actually worked was this:



 proposition four_points_sufficient: "affine_plane plmeets"
unfolding affine_plane_def
apply (intro conjI)
subgoal using four_points_a1 by blast


It's not clear to me how "It is possible to do this [i.e., apply conjI twice] within an Isar proof," but perhaps I don't now need to know.







share|improve this answer














share|improve this answer



share|improve this answer








answered 2 days ago


























community wiki





John








  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday













  • 1





    "It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

    – xanonec
    yesterday






  • 1





    Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

    – xanonec
    yesterday








1




1





"It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

– xanonec
yesterday





"It is possible to do this [i.e., apply conjI twice] within an Isar proof,". It is possible to use a comma-separated list of methods (sequential composition, also see 6.4 in the reference manual) both in Isar and apply scripts, e.g lemma "True" proof fix A B :: "'a ⇒ bool" have "∀x y. A x ∧ B y ⟶ B y ∧ A x" by (rule allI, rule allI, simp) ....

– xanonec
yesterday




1




1





Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

– xanonec
yesterday






Also, there is nothing that could prevent one from using apply style script inside a proof block. To the best of my knowledge, it is not considered to be a very good practice, but it is not considered to be a particularly bad practice either (see "Do not switch from apply into proof in the middle of a proof" in https://proofcraft.org/blog/isabelle-style.html).

– xanonec
yesterday




















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