- #1
Barioth
- 49
- 0
Hi
here is the problem I'm working on,
Let $A(x_1)$ be a well formed formula of a language $L$ in which $x_1$ is free, let $a_1$ an invidual constant of $L$, Show that the formula $A(a_1)\rightarrow(\exists x_1)A(x_1)$ is a theorem of $K_L$
on this link, the first slide as the axiom of $K_L$
http://www.idi.ntnu.no/emner/tdt4135/handouts/slides-9.pdf
I wanted to use the following proof:
$A(a_1)\rightarrow(\exists x_1)A(x_1)$
is equivalent logic to
$(\exists x_1)(A(a_1)\rightarrow A(x_1))$
wich is equivalent logic to
$(\forall x_1)A(a_1)\rightarrow A(x_1)$
then trying to conclude using (K5) from the pdf link,
but I just realized that since $x_1$ is free in A, I cannot move the quantifier like I did.
here is the problem I'm working on,
Let $A(x_1)$ be a well formed formula of a language $L$ in which $x_1$ is free, let $a_1$ an invidual constant of $L$, Show that the formula $A(a_1)\rightarrow(\exists x_1)A(x_1)$ is a theorem of $K_L$
on this link, the first slide as the axiom of $K_L$
http://www.idi.ntnu.no/emner/tdt4135/handouts/slides-9.pdf
I wanted to use the following proof:
$A(a_1)\rightarrow(\exists x_1)A(x_1)$
is equivalent logic to
$(\exists x_1)(A(a_1)\rightarrow A(x_1))$
wich is equivalent logic to
$(\forall x_1)A(a_1)\rightarrow A(x_1)$
then trying to conclude using (K5) from the pdf link,
but I just realized that since $x_1$ is free in A, I cannot move the quantifier like I did.