Mixing
Prove that ∼∼P ⊢P (Sider)
$begingroup$
2.4(c)
- ∼∼P premise .
- ∼∼P →(∼P →∼∼P) PL1 .
- ∼P→∼∼P 1,2,MP .
- (∼P →∼∼P)→((∼P →∼P)→P) PL3 .
- (∼P→∼P)→P 3,4,MP .
- ∼P→∼P premise 2.4(b) .
- P 5,6,MP .
This is the answer, what I don't understand is line 6 when it says ~P->~P is a premise according to 2.4(b).
Here is the answer to 2.4(b)
We need to prove that ⊢(∼P →P)→P
- ∼P →((∼P →∼P)→∼P)→ etc. PL2
- ∼P →((∼P →∼P)→∼P) PL1
- (∼P →(∼P →∼P))→(∼P →∼P) 1,2,MP
- ∼P→(∼P→∼P) PL1
- ∼P→∼P 3,4 MP
- (∼P →∼P)→(∼P →P)→P PL3
- (∼P→P)→P 5,6,MP
So, I see that in line 5 that it's the same as in line 6. What I don't understand is how you're able to cite a completely different proof. If I didn't have 2.4(b), how would I be able to continue the proof from line 5 in 2.4(c)?
The axioms are defined as:
- P->(Q->P) PL1
- (P->(Q->X))->(P->Q)->(P->X)) PL2
- (~Q->~P)->((~Q->P)->Q) PL3
logic
asked Feb 1 at 17:54
ineedhelpineedhelp
112
$endgroup$
|
show 4 more comments
$begingroup$
2.4(c)
- ∼∼P premise .
- ∼∼P →(∼P →∼∼P) PL1 .
- ∼P→∼∼P 1,2,MP .
- (∼P →∼∼P)→((∼P →∼P)→P) PL3 .
- (∼P→∼P)→P 3,4,MP .
- ∼P→∼P premise 2.4(b) .
- P 5,6,MP .
This is the answer, what I don't understand is line 6 when it says ~P->~P is a premise according to 2.4(b).
Here is the answer to 2.4(b)
We need to prove that ⊢(∼P →P)→P
- ∼P →((∼P →∼P)→∼P)→ etc. PL2
- ∼P →((∼P →∼P)→∼P) PL1
- (∼P →(∼P →∼P))→(∼P →∼P) 1,2,MP
- ∼P→(∼P→∼P) PL1
- ∼P→∼P 3,4 MP
- (∼P →∼P)→(∼P →P)→P PL3
- (∼P→P)→P 5,6,MP
So, I see that in line 5 that it's the same as in line 6. What I don't understand is how you're able to cite a completely different proof. If I didn't have 2.4(b), how would I be able to continue the proof from line 5 in 2.4(c)?
The axioms are defined as:
- P->(Q->P) PL1
- (P->(Q->X))->(P->Q)->(P->X)) PL2
- (~Q->~P)->((~Q->P)->Q) PL3
logic
asked Feb 1 at 17:54
ineedhelpineedhelp
112
$endgroup$
$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
1
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19
|
show 4 more comments
$begingroup$
2.4(c)
- ∼∼P premise .
- ∼∼P →(∼P →∼∼P) PL1 .
- ∼P→∼∼P 1,2,MP .
- (∼P →∼∼P)→((∼P →∼P)→P) PL3 .
- (∼P→∼P)→P 3,4,MP .
- ∼P→∼P premise 2.4(b) .
- P 5,6,MP .
This is the answer, what I don't understand is line 6 when it says ~P->~P is a premise according to 2.4(b).
Here is the answer to 2.4(b)
We need to prove that ⊢(∼P →P)→P
- ∼P →((∼P →∼P)→∼P)→ etc. PL2
- ∼P →((∼P →∼P)→∼P) PL1
- (∼P →(∼P →∼P))→(∼P →∼P) 1,2,MP
- ∼P→(∼P→∼P) PL1
- ∼P→∼P 3,4 MP
- (∼P →∼P)→(∼P →P)→P PL3
- (∼P→P)→P 5,6,MP
So, I see that in line 5 that it's the same as in line 6. What I don't understand is how you're able to cite a completely different proof. If I didn't have 2.4(b), how would I be able to continue the proof from line 5 in 2.4(c)?
The axioms are defined as:
- P->(Q->P) PL1
- (P->(Q->X))->(P->Q)->(P->X)) PL2
- (~Q->~P)->((~Q->P)->Q) PL3
logic
asked Feb 1 at 17:54
ineedhelpineedhelp
112
$endgroup$
2.4(c)
- ∼∼P premise .
- ∼∼P →(∼P →∼∼P) PL1 .
- ∼P→∼∼P 1,2,MP .
- (∼P →∼∼P)→((∼P →∼P)→P) PL3 .
- (∼P→∼P)→P 3,4,MP .
- ∼P→∼P premise 2.4(b) .
- P 5,6,MP .
This is the answer, what I don't understand is line 6 when it says ~P->~P is a premise according to 2.4(b).
Here is the answer to 2.4(b)
We need to prove that ⊢(∼P →P)→P
- ∼P →((∼P →∼P)→∼P)→ etc. PL2
- ∼P →((∼P →∼P)→∼P) PL1
- (∼P →(∼P →∼P))→(∼P →∼P) 1,2,MP
- ∼P→(∼P→∼P) PL1
- ∼P→∼P 3,4 MP
- (∼P →∼P)→(∼P →P)→P PL3
- (∼P→P)→P 5,6,MP
So, I see that in line 5 that it's the same as in line 6. What I don't understand is how you're able to cite a completely different proof. If I didn't have 2.4(b), how would I be able to continue the proof from line 5 in 2.4(c)?
The axioms are defined as:
- P->(Q->P) PL1
- (P->(Q->X))->(P->Q)->(P->X)) PL2
- (~Q->~P)->((~Q->P)->Q) PL3
logic
logic
asked Feb 1 at 17:54
ineedhelpineedhelp
112
asked Feb 1 at 17:54
ineedhelpineedhelp
112
edited Feb 1 at 22:08
ineedhelp
asked Feb 1 at 17:54
ineedhelpineedhelp
112
asked Feb 1 at 17:54
ineedhelpineedhelp
112
asked Feb 1 at 17:54
ineedhelpineedhelp
112
112
$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
1
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19
|
show 4 more comments
$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
1
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19
$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
1
1
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First, line 5) from 2.4c is nothing more than MP applied to lines 3) and 4), as indicated. That is, MP is defined as:
$varphi to psi$
$varphi$
$therefore psi$
In this partiuclar case, line 3) is the $varphi$. That is, use $varphi= neg P to neg neg P$
Line 5) is the $psi$. That is: $psi= (neg P to neg P) to P$
And line 4) is the $varphi to psi$, i.e. $(neg P to neg neg P) to ((neg P to neg P) to P)$
Second, if you didn't have the proof for 2.4b, then after line 5) you would simply repeat the very 7 lines from the proof of 2.4b, so these would become lines 6 through 12, where line 12 would be $(neg P to P) to P$
And then what is now line 7 becomes line 13
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
$endgroup$
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First, line 5) from 2.4c is nothing more than MP applied to lines 3) and 4), as indicated. That is, MP is defined as:
$varphi to psi$
$varphi$
$therefore psi$
In this partiuclar case, line 3) is the $varphi$. That is, use $varphi= neg P to neg neg P$
Line 5) is the $psi$. That is: $psi= (neg P to neg P) to P$
And line 4) is the $varphi to psi$, i.e. $(neg P to neg neg P) to ((neg P to neg P) to P)$
Second, if you didn't have the proof for 2.4b, then after line 5) you would simply repeat the very 7 lines from the proof of 2.4b, so these would become lines 6 through 12, where line 12 would be $(neg P to P) to P$
And then what is now line 7 becomes line 13
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
$endgroup$
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
add a comment |
$begingroup$
First, line 5) from 2.4c is nothing more than MP applied to lines 3) and 4), as indicated. That is, MP is defined as:
$varphi to psi$
$varphi$
$therefore psi$
In this partiuclar case, line 3) is the $varphi$. That is, use $varphi= neg P to neg neg P$
Line 5) is the $psi$. That is: $psi= (neg P to neg P) to P$
And line 4) is the $varphi to psi$, i.e. $(neg P to neg neg P) to ((neg P to neg P) to P)$
Second, if you didn't have the proof for 2.4b, then after line 5) you would simply repeat the very 7 lines from the proof of 2.4b, so these would become lines 6 through 12, where line 12 would be $(neg P to P) to P$
And then what is now line 7 becomes line 13
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
$endgroup$
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
add a comment |
$begingroup$
First, line 5) from 2.4c is nothing more than MP applied to lines 3) and 4), as indicated. That is, MP is defined as:
$varphi to psi$
$varphi$
$therefore psi$
In this partiuclar case, line 3) is the $varphi$. That is, use $varphi= neg P to neg neg P$
Line 5) is the $psi$. That is: $psi= (neg P to neg P) to P$
And line 4) is the $varphi to psi$, i.e. $(neg P to neg neg P) to ((neg P to neg P) to P)$
Second, if you didn't have the proof for 2.4b, then after line 5) you would simply repeat the very 7 lines from the proof of 2.4b, so these would become lines 6 through 12, where line 12 would be $(neg P to P) to P$
And then what is now line 7 becomes line 13
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
$endgroup$
First, line 5) from 2.4c is nothing more than MP applied to lines 3) and 4), as indicated. That is, MP is defined as:
$varphi to psi$
$varphi$
$therefore psi$
In this partiuclar case, line 3) is the $varphi$. That is, use $varphi= neg P to neg neg P$
Line 5) is the $psi$. That is: $psi= (neg P to neg P) to P$
And line 4) is the $varphi to psi$, i.e. $(neg P to neg neg P) to ((neg P to neg P) to P)$
Second, if you didn't have the proof for 2.4b, then after line 5) you would simply repeat the very 7 lines from the proof of 2.4b, so these would become lines 6 through 12, where line 12 would be $(neg P to P) to P$
And then what is now line 7 becomes line 13
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
answered Feb 1 at 23:02
Bram28Bram28
64.4k44793
64.4k44793
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
add a comment |
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
$begingroup$
thank you @Bram28 ! i really appreciate it
$endgroup$
– ineedhelp
Feb 2 at 0:40
add a comment |
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$begingroup$
How is your axiom PL3 defined?
$endgroup$
– Henning Makholm
Feb 1 at 18:02
$begingroup$
Hi @Henning Makholm, thanks for replying I added the edits to the post
$endgroup$
– ineedhelp
Feb 1 at 18:11
$begingroup$
Are you really sure the last $P$ in your PL3 should not be a $Q$?
$endgroup$
– Henning Makholm
Feb 1 at 18:28
$begingroup$
@HenningMakholm you're right! it's supposed to be a Q
$endgroup$
– ineedhelp
Feb 1 at 18:45
1
$begingroup$
You say "Why is it ((P->~P)->P)?" ... but I don't see that statement anywhere ...
$endgroup$
– Bram28
Feb 1 at 19:19