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
  • Nope, that is not valid. Are you missing a 'not' before that first P? May 1, 2019 at 23:35
  • Is it not valid because there's no (NOT)P in the premises? Since it's not known? No, it's not missing. The first premise is P --> Q. May 1, 2019 at 23:42
  • Then you cannot conclude ¬P → (Q ˄ R) from those premises, although you may conclude P → (Q ˄ R) May 1, 2019 at 23:47
  • The easiest answer is this: you cannot have a negative conclusion with all positive premises. The NOT is considered a Negative. Any premise with NOT appearing would be deemed a negative premise. However you say there are no negative premises but the is a NOT in the conclusion. The NOT in the conclusion would make the conclusion negative. There should be a red flag if the conclusion is negative and all premises are positive.
    – Logikal
    May 3, 2019 at 14:03

3 Answers 3


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

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.

  • No, there's no typo. I imagined that this proof wasn't valid simply because I coulnd't conclude (NOT)P from anywhere. So, when I replace the letters with cube(a) , cube(b), why is it showing, in the world, that the proof is valid? Should I choose names that I know will prove it's invalid? May 1, 2019 at 23:53
  • An argument is (semantically) valid when the conclusion can not be false in all interpretations that hold every premise to be true. However, finding that there may be some interpretation where the conclusion and premises are all true is not enough to demonstrate validity. Jun 14, 2019 at 15:12

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.

  • That makes sense. So the reason that the proof is invalid, is because out of all the results, there's only 1 that is false? In a valid proof, they should all be true? May 1, 2019 at 23:58
  • That's right. If there were only T values then the statement would be a tautology based on the truth table and a derivation or proof could be found that would be valid. May 2, 2019 at 0:01

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)counterexample world (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.

  • Hey thanks for the help! I do have to use Tarski's World as a tool. I tried myself to make the premises true and the conclusion false -- but it didn't work. I guess you did it differently! I assigned Cube(a) to P, Cube(b) to Q, Large(a) to R, and Large(b) to S, and both the premises and the goal was valid. Odd. May 2, 2019 at 17:20
  • So the reason why that happened is because "a" really was a cube. What I don't understand is how Large(a) when "a" is not a cube -- I changed it to a Dodecahedron and yet the premise : Cube(a) --> Large(a) is still valid, even when Dodec(a). How does that happen? May 2, 2019 at 17:30
  • 1
    So I tried it out with your predicates and if you make a world with two large dodecs (or tets) being a and b, then your premises will come out true and your conclusion will come out false, making it invalid.
    – Evan
    May 2, 2019 at 20:35

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