Thread: Predicate calculus View Single Post
 P: 19 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