- #1
solakis
- 19
- 0
Given the following :
1)[itex]\forall x\forall y\forall z G(F(F(x,y),z),F(x,F(y,z)))[/itex]
2)[itex]\forall xG(F(x,c),x)[/itex]
3)[itex]\forall x\exists yG(F(x,y),c)[/itex]
4)[itex]\forall x\forall yG(F(x,y),F(y,x))[/itex].
5) [itex]\forall x\forall y\forall z ( G(x,y)\wedge G(x,z)\Longrightarrow G(y,z))[/itex]
Where G is a two place predicate symbol. F ,is a two place term symbol and c is a constant.
Prove :[itex]\exists! y\forall xG(F(x,y),x)[/itex]
[itex]\exists ! y[/itex] means : there exists a unique y
1)[itex]\forall x\forall y\forall z G(F(F(x,y),z),F(x,F(y,z)))[/itex]
2)[itex]\forall xG(F(x,c),x)[/itex]
3)[itex]\forall x\exists yG(F(x,y),c)[/itex]
4)[itex]\forall x\forall yG(F(x,y),F(y,x))[/itex].
5) [itex]\forall x\forall y\forall z ( G(x,y)\wedge G(x,z)\Longrightarrow G(y,z))[/itex]
Where G is a two place predicate symbol. F ,is a two place term symbol and c is a constant.
Prove :[itex]\exists! y\forall xG(F(x,y),x)[/itex]
[itex]\exists ! y[/itex] means : there exists a unique y