# 2 simple Formal Fitch Proofs

I'm having difficulty proving these. They seem obvious, but I can't figure how to set up formal proofs for them. Could anyone give me clues on how to start them?

1. ¬(P∧¬Q) from the premise P→Q;

2. ¬Q→(R→P) from the premise P∨Q.

Thanks

• I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags. – Frank Hubeny Oct 8 '18 at 2:15
• it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using? – virmaior Oct 8 '18 at 3:04
• As an aside, note that the first of these will necessarily require use of the law of excluded middle (or proof by contradiction) since it is not constructively true but it is a truth-table tautology. The second is constructively true: you can prove it without contradiction. – Patrick Stevens Oct 8 '18 at 6:25

¬(P∧¬Q) from the premise P→Q

Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.

1. P→Q

this means that either P is false or Q is true based on the truth table that defines →.

Using a truth table, we can then check to see if the two are equivalent:

`````` P   |   Q   |   P → Q |  ¬Q  |  P ∧ ¬Q  | ¬(P∧¬Q)
T       T         T       F       F          T
T       F         F       T       T          F
F       T         T       F       F          T
F       F         T       T       F          T
``````

So we've shown they are equivalent via truth tables.

Now, how do we move that into the realm of natural deduction?

The two basic options are:

1. use inference rules
2. do a sub proof -- either to get a contradiction or supply a conditional proof

Since our conclusion has a negation, let's go with a subproof by contradiciton.

1. P→Q A
2. | P∧¬Q A
3. | P ∧E 2
4. | Q MP 1,3
5. | ¬Q ∧E 2
6. | ⊥ contradiction on 4,5
7. ¬(P∧¬Q) RAA 2-6 References

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

(references copied from Frank Hubeny)

I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.

Here are two proofs using Klement's proof checker. The rules you may have to use may be different. The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).

To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".

For the other one, try something like this: The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).

Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".

There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.

At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.

You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

• Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful! – 35308 Oct 8 '18 at 4:49

¬(P∧¬Q) from the premise P→Q;

To derive a negation of a proposition from a conditional statement, assume the proposition to hopefully derive a contradiction by eliminating the conditional, so that the negation may be introduced.

Given the premise of P→Q , assume (P∧¬Q), derive a contradiction, then introduce a negation. Thereby deriving ¬(P∧¬Q) from P→Q.

¬Q→(R→P) from the premise P∨Q.

To derive a condtional proposition from a disjunction, use a proof by cases where the subproofs are conditional proofs.

Given the premise P∨Q, seek to eliminate the disjunction. Assume each case (P, Q) in turn to derive (¬Q→(R→P) (hint: restatement and explosion). Thereby deriving (¬Q→(R→P) from P∨Q.