I'm having a hard time proving that this is a valid argument

$Premise 1: (Ǝx)Kx→(\forall x)(Lx→Mx)$

$Premise 2: Kc • Lc$

$Conclusion: Mc$

I am getting confused with all the existential/universal instantiation rules.

First I did: $Kc\to\forall x(Lx\to Mx)$ by existential instantiation

then $Kc\to Lc\to Mc$ by universal instantiation

therefore $Mc$

Is there anywhere where I'm supposed to infer modus ponens since I've already been given Kc and Lc in premise 2??

  • $\begingroup$ Yes, probably. The details will depend on your inference rules. Your first step is wrong, though.Existential instantiation is for existential statements and the first premise is a conditional statement, not an existential one. $\endgroup$ – Git Gud Sep 5 '15 at 17:00
  • $\begingroup$ But what if I break the conditinal into givens and a goal, the given will be an existential statement, i can then use existential instantiation?? $\endgroup$ – pkjag Sep 5 '15 at 17:06
  • $\begingroup$ Check Mauro's hint. From $Kc$ you get $(\exists x)Kx$ and then you can use modus ponens (for the first time). You'll have to use it a second time. $\endgroup$ – Git Gud Sep 5 '15 at 17:10

Hint : assuming you are using Natural Deduction ...

From Premise-2 : $Kc \land Lc$ you have to derive $Kc$ (by $\land$-elimination), followed by $\exists x Kx$ (by $\exists$-introduction).

In this way, you can use $\to$-elimination (i.e. modus ponens) with Premise-1 and derive : $(∀x)(Lx \to Mx)$.

Now you have to use $\forall$-elimination (i.e. universal instantiation) with $c$ to get : $Lc \to Mc$.


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