## predicate calculus

Given the following :

1)$\forall x\forall y\forall z G(F(F(x,y),z),F(x,F(y,z)))$

2)$\forall xG(F(x,c),x)$

3)$\forall x\exists yG(F(x,y),c)$

4)$\forall x\forall yG(F(x,y),F(y,x))$.

5) $\forall x\forall y\forall z ( G(x,y)\wedge G(x,z)\Longrightarrow G(y,z))$

Where G is a two place predicate symbol. F ,is a two place term symbol and c is a constant.

Prove :$\exists! y\forall xG(F(x,y),x)$

$\exists ! y$ means : there exists a unique y
 PhysOrg.com science news on PhysOrg.com >> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens>> Google eyes emerging markets networks
 Is this a homework question, or just for fun, or what? And I am assuming that E! x ph <-> E. y A. x ( x = y <-> ph ), right?
 This is a problem given to me by a friend ,that i could not solve out. This two place predicate and term is very confusing.

## predicate calculus

Is "c" a constant or a variable? Because if it is a variable, then you can prove a contradiction given the conclusion and given that there are at least 2 distinct values of x.

We could prove that A. y A. x G ( F ( x , y ) x ) given 2, which contradicts the conclusion.
 I have mention it already in my opening post that c is a constant
 Well, existence is straight forward: From 2) and c $$\exists c \wedge \forall x G(F(x,c),x) \Rightarrow \exists y \forall x G(F(x,y),x)$$ uniqueness is left as an exercise: $$\forall u(\forall x G(F(x,u),x) \Rightarrow u=c)$$
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus When you take G the equality, then your list of axioms is that of a commutative group. The thing you need to prove is that the identity element is unique. Take two identity's c and c', then 3) G(F(c',c),c') and 3) G(F(c,c'),c) and 4) G(F(c,c'),F(c',c)) So by (5), we get that G(c,c') But that doesn't give equality, however...
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus OK, what about this counterexample: Take $\mathbb{Z}_0$ as universe. Take $G(x,y)~\text{if and only if}~\frac{x}{y}\geq 0$ and F(x,y)=x*y and c=1. Then y=1 and y=2 both satisfy the hypothesis.