# Rules of Inference argument has multiple steps?

I need to write the rules of inference and explain which rules of inference are used for each step. This looks like it's just Modus Ponens from what I wrote the equations as, am I missing some steps or is this really that simple?

“There is someone in this class who has been to France. Everyone who goes to France visits the Louvre. Therefore, someone in this class has visited the Louvre.”

C(x) “x is in this class”, F(x) “x has been to France”, L(x) “x visited the Louvre”
∃x(C(x) ∧ F(x))
∀x(F(x)→L(x))
∃x(C(x) ∧ L(x))

Your rendering into QL, I am afraid, is hopelessly wrong! It is not even well-formed (count brackets!), and anyway you need to represent two premisses and a conclusion, so three separate QL wffs. It should be

$\exists x(Sx \land Fx)$, $\forall x(Fx \to Lx)$, so $\exists x(Sx \land Lx)$

For particularly good books on translating into QL, Paul Teller's Modern Logic Primer is freely available at http://tellerprimer.ucdavis.edu/ Or there's my Introduction to Formal Logic!

• Yeah I realized that as I looked back through the book. I was typing up the answer (from what I understand) hopefully it helps others. Jul 6, 2012 at 6:48

I ended up figuring it out and I think this is 100% correct for the answer. My logic could be incorrect thought since I'm still learning this stuff.

C(x) “x is in this class”, F(x) “x has been to France”, L(x) “x visited the Louvre”
1. ∃x(C(x) ∧ F(x))
2. C(x) ∧ F(x) existential instantiation
3. F(x) simplification from 2
4. ∀x(F(x)→L(x))
5. F(x)→L(x) universal instantiation
6. L(x) modus ponens from 3 and 5
7. C(x) simplification from 2
8. C(x) ∧ L(x) conjunction from 6 and 7
9. ∃x(C(x) ∧ L(x)) existential generalization

• That looks pretty good. If I were you I would distinguish between the bound variable $x$ in steps 1, 4, and 9, and the $x$ in steps 2, 4, 5–8 that represents a particular individual for whom $C(x)\land F(x)$ happens to be true. Perhaps use $k$ for this individual, rather than $x$.
– MJD
Jul 8, 2012 at 0:57