Fitch Proof Question

Question

I'm having trouble with a proof and I'm not sure if it's valid or not. If it appears to be invalid, we are supposed to assign names to the letters in the proof and check it in a World, but when I do it still checks out.

When I try to prove it in Fitch, I get all the way to the bottom, with (Q^R) as the last step of the subproof and I'm trying to apply -->Intro to join (NOT)P --> (Q ^ R) but it's not checking out because it says that it needs a support step to be cited.

P --> Q                Premise
R ^ S                  Premise
------------------
(NOT)P --> (Q ^ R)     Goal

Answer 1

P → Q, R ˄ S |- ¬P → (Q ˄ R) is not syntactically valid.

An assumption of ¬P will not allow you to eliminate the conditional in P → Q to derive Q .

Is there a typo?

P → Q, R ˄ S |- P → (Q ˄ R) is valid.

 1|  P → Q
 2|_ R ˄ S
 3|  R               ˄E, 2
 4|  |_ P
 5|  |  Q            →E, 1,4
 6|  |  Q ˄ R        ˄I, 3,5
 7|  P → (Q ˄ R)    →I, 4-6

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For an argument to be semantically valid, the conclusion must be demonstrably true in all interpretations where the premises are -- it is not enough to find just one. A proof is semantically invalid when the exists some interpretation where all the premises are true, but the conclusion is false. One such counter example is enough to disprove an argument.

In the interpretation of {P=f,Q=f, R=t, S=t}, the premises P → Q, R ˄ S, are both true, but ¬P → (Q ˄ R) is false.

Answer 2

By using a truth table generator one can show that one of the set of valuations for the sentence letters leads to the result being false.

To see this, use this input ((P=>Q)&&(R&&S))=>(~P=>(Q&&R)) in the Stanford Truth Table Tool. When P and Q are false and R and S are true then the conditional with the conjunction of the premises as antecedent and the goal as consequent is false.

This would mean that one could not derive the result.

Answer 3

It's invalid. When there can be a world in which the premises are true and the conclusion is false, the argument is invalid. So if you wanted to use Tarsky's World as tool, you would need to use block language. So lets make: P = Tet(a) Q = Small(a) R = Cube(b) S = Large(b) (these are just random predicates that don't change the meaning of the argument) If there are two cubes, one being large(b) and one being medium(a), then both premises would be true and the conclusion (given that a is not small in this world) would be false.