Re: how to prove E!x(Fx & Gx) -> Ax(Fx-> Gx)

Am Thu, 4 Sep 2008 23:29:20 +0200 schrieb Balthasar:


... with other words,

E!xFx & Ex(Fx & Gx) -> Ax(Fx -> Gx).

Now this one obviously is valid. Non-formal proof:

Assume E!xFx & Ex(Fx & Gx). Hence E!xFx and Ex(Fx & Gx). Since E!xFx there
is exactly one F. Assume that a is this F, hence Fa. Now from Ex(Fx & Gx)
we get that there is an x such that Fx & Gx. But since there is exactly one
F, namely a, we get: Fa & Ga. Now assume Fb for an arbitrary b. Then either
b = a or b =/= a. If b = a, then Fb & Gb (since Fa & Ga), hence Fb -> Gb
(propositional logic). If b =/= a, then ~Fb (since a is the only F), hence
Fb -> Gb (propositional logic). Hence in any case Fb -> Gb. Since b was
arbitrary, this means for any x: Fx -> Gx. With other words, Ax(Fx -> Gx).

Too complicated. We now have a simpler proof:

Assume E!xFx & Ex(Fx & Gx). Hence E!xFx and Ex(Fx & Gx). Assume Fa & Ga.
Hence Fa and Ga. Now assume for an arbitrary b: Fb. Then a = b (since there
is exactly one F, and Fa). Hence Gb (from Ga and a = b). Hence Fb -> Gb.
Since b was arbitrary, this means for any x: Fx -> Gx. With other words,
Ax(Fx -> Gx). []


B.
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