# Fitch Proof - Logic LPL 6.31

I am trying to complete the following proof in Fitch but am completely clueless on how to approach it.

Any help would be appreciated! Thanks

• The premises do not entail the goal. Please check for miscopying. Oct 7, 2019 at 10:07
• Ah, "you should assess whether the argument is valid." And it isn't. See if you can satisfy the three premises while making the conclusion false. Oct 7, 2019 at 11:52
• Thank you! I forgot to consider the validity of the argument. Oct 7, 2019 at 18:06

As Mauro ALLEGRANZA notes in an answer, it is possible that the argument is not valid. This answer will provide two ways to check if that is the case.

First note that although the argument uses predicates of an object b there are no existential or universal quantifiers. We may think of each of these predicates as propositions. This will allow us to simplify the symbolization as follows:

• Dodec(b) as D
• Cube(b) as C
• Small(b) as S
• Medium(b) as M

Then the premises can be rewritten as

1. D v C
2. S v M
3. ~(S & C)

and the goal as

M & D.

Here are the two ways to check if that is not a valid argument:

One can conjoin the premises, connect this conjunction to the goal with a conditional, and enter that resulting proposition into a truth table generator to arrive at the following:

If this were valid all of the rows in the truth table would show 'T' under the conditional column in red. There are two rows which do not and so the argument is not valid.

It is worth noting why those two rows stand out: in both of them the goal is false while the conjoined premises are true.

The second way is to use a tree proof generator as follows:

The generator returns a countermodel. It found this by assuming the goal was false and looking for valuations that made the premises true by attempting to construct a tree proof. The branches of the tree that did not close provided the generator with the desired valuations for the countermodel. It returned the valuations from one of those branches.

Note that the valuation obtained is the sixth row in the truth table above. One only needs one such valuation to show that the argument is invalid.

Either of these methods could be used to find a countermodel. One could manually find this countermodel this as well. Look for valuations that make M & D false. Use each of those valuations and try to find valuations for the other propositional variables that make the premises true. If one can do that one has found a countermodel or counterexample showing the argument is invalid.

Tree Proof Generator. https://www.umsu.de/trees/

Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html

Hint

See page 174 :

"In each of the following exercises, you should assess whether the argument is valid. "

Use truth table with the following truth-assignment : v(M)=TRUE and v(D)=FALSE.

I had this picture in my files and I believe it is correct.

• Well, that is valid, however your third premise is not the equivalent to the one in the original question. May 18, 2020 at 6:18