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Supervision of formal proof of cut rule

1

0

$begingroup$

The Cut rule states that from the formulas $(F to G)$ and $(G to H)$
we can derive the formula $(F to H)$.

My task is to give a formal proof for this rule. I'm not sure if my proof is formally and logically correct, so I'm asking for supervision.

$$
begin{alignat*}{}
1.&Gammavdash(Fto G)&text{Premise 1}\
2.&Gammavdash(Gto H)&text{Premise 2}\
3.&Gamma,cup{F}vdash F&text{Assume $F$}\
4.&Gamma,cup{F}vdash(Fto G)&text{Apply monotonicity to 1}\
5.&Gamma,cup,{F}vdash G&text{Apply $to$ elimination to 3 and 4}\
6.&Gamma,cup,{F}vdash(Gto H)&text{Apply monotonicity to 2}\
7.&Gamma,cup,{F}vdash H&text{Apply $to$ elimination to 5 and 6}\
8.&Gammavdash(Fto H)&text{Apply $to$ introduction to 7}\
end{alignat*}
$$

Note: This is exercise 1.9 from 1st edition of A First Course in Logic.

proof-verification logic

share|cite|improve this question

edited Mar 25 at 20:01

user1518183

asked Mar 25 at 16:46

user1518183user1518183

1716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that step 3 is justified by Monotonicty rule and $to$-elimination.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 25 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MauroALLEGRANZA Thank you, I'll fix it.
  $endgroup$
  – user1518183
  Mar 25 at 16:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And you need an additional step 6. $Gamma vdash (F to H)$ by $to$-introduction.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 25 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is "reversed $to$ introduction"?
  $endgroup$
  – Henning Makholm
  Mar 25 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HenningMakholm It's not a real rule and should be rather qoted. It logically make sense to me, but formally it will be maybe sequence of assumption/monotonicity and $to$ elimination.
  $endgroup$
  – user1518183
  Mar 25 at 17:11
  
  
  

  
  
  
  

 | 
show 1 more comment

1

0

$begingroup$

The Cut rule states that from the formulas $(F to G)$ and $(G to H)$
we can derive the formula $(F to H)$.

My task is to give a formal proof for this rule. I'm not sure if my proof is formally and logically correct, so I'm asking for supervision.

$$
begin{alignat*}{}
1.&Gammavdash(Fto G)&text{Premise 1}\
2.&Gammavdash(Gto H)&text{Premise 2}\
3.&Gamma,cup{F}vdash F&text{Assume $F$}\
4.&Gamma,cup{F}vdash(Fto G)&text{Apply monotonicity to 1}\
5.&Gamma,cup,{F}vdash G&text{Apply $to$ elimination to 3 and 4}\
6.&Gamma,cup,{F}vdash(Gto H)&text{Apply monotonicity to 2}\
7.&Gamma,cup,{F}vdash H&text{Apply $to$ elimination to 5 and 6}\
8.&Gammavdash(Fto H)&text{Apply $to$ introduction to 7}\
end{alignat*}
$$

Note: This is exercise 1.9 from 1st edition of A First Course in Logic.

proof-verification logic

share|cite|improve this question

edited Mar 25 at 20:01

user1518183

asked Mar 25 at 16:46

user1518183user1518183

1716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that step 3 is justified by Monotonicty rule and $to$-elimination.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 25 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MauroALLEGRANZA Thank you, I'll fix it.
  $endgroup$
  – user1518183
  Mar 25 at 16:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And you need an additional step 6. $Gamma vdash (F to H)$ by $to$-introduction.
  $endgroup$
  – Mauro ALLEGRANZA
  Mar 25 at 16:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is "reversed $to$ introduction"?
  $endgroup$
  – Henning Makholm
  Mar 25 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @HenningMakholm It's not a real rule and should be rather qoted. It logically make sense to me, but formally it will be maybe sequence of assumption/monotonicity and $to$ elimination.
  $endgroup$
  – user1518183
  Mar 25 at 17:11
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

The Cut rule states that from the formulas $(F to G)$ and $(G to H)$
we can derive the formula $(F to H)$.

My task is to give a formal proof for this rule. I'm not sure if my proof is formally and logically correct, so I'm asking for supervision.

$$
begin{alignat*}{}
1.&Gammavdash(Fto G)&text{Premise 1}\
2.&Gammavdash(Gto H)&text{Premise 2}\
3.&Gamma,cup{F}vdash F&text{Assume $F$}\
4.&Gamma,cup{F}vdash(Fto G)&text{Apply monotonicity to 1}\
5.&Gamma,cup,{F}vdash G&text{Apply $to$ elimination to 3 and 4}\
6.&Gamma,cup,{F}vdash(Gto H)&text{Apply monotonicity to 2}\
7.&Gamma,cup,{F}vdash H&text{Apply $to$ elimination to 5 and 6}\
8.&Gammavdash(Fto H)&text{Apply $to$ introduction to 7}\
end{alignat*}
$$

Note: This is exercise 1.9 from 1st edition of A First Course in Logic.

proof-verification logic

share|cite|improve this question

edited Mar 25 at 20:01

user1518183

asked Mar 25 at 16:46

user1518183user1518183

1716

$endgroup$

share|cite|improve this question

edited Mar 25 at 20:01

user1518183

asked Mar 25 at 16:46

user1518183user1518183

1716

share|cite|improve this question

edited Mar 25 at 20:01

user1518183

asked Mar 25 at 16:46

user1518183user1518183

1716

asked Mar 25 at 16:46

user1518183user1518183

1716

asked Mar 25 at 16:46

user1518183user1518183

1716