Can You Prove These Hypotheses in Predicate Calculus?

[stauros]

Given :

a)

1) c is a constant

2) P and K are one place operation symbols

3) G and H are a two place predicate symbols

b)

The following hypothesis

1)for all A { G(A,A) }

2) for all A,B { H(A,c) =>( G[P(A),B] <=> ( G[K(B),A] and H(B,c)))}

Then prove :

1) for all A { H(A,c) => G[K(P(A),A] }

2) for all A { H(A,c) => H( P(A),c) }

 

[dcpo]

How you formally work this through depends on what system you're asked to work with, but a strategy would be to let B=P(A).

 

[stauros]

Thanks dcpo .So i put B=P(A) in 2 and i get :

{ H(A,c) =>( G[P(A),P(A)] <=> ( G[K(P(A)),A] and H(P(A),c)))}
is that correct?

The proof is in predicate calculus ,so every line of the proof has to be accounted for and justified

 

[dcpo]

Well, what you need to say depends on how formal you have to be, and on what deduction system you have to use, but what you have is the basis for a rigorous 'everyday' proof, so long as you make explicit the role of the $\forall A(G(A,A))$ hypothesis. If you haven't been given an explicit formal deduction system to work with that should be enough.

 

[stauros]

The proof as i said is an ordinary proof in predicate calculus with the usal 4 general laws i.e

1) Universal Elimination
2) Universal Introduction
3) Existential Elimination
4) Existential Introduction

Plus the rules of statement calculus

 

[xxxx0xxxx]

Given :

a)
1) c is a constant
2) P and K are one place operation symbols
3) G and H are a two place predicate symbols

b)
The following hypothesis
1)for all A { G(A,A) }
2) for all A,B { H(A,c) =>( G[P(A),B] <=> ( G[K(B),A] and H(B,c)))}

Then prove :

1) for all A { H(A,c) => G[K(P(A),A] }
2) for all A { H(A,c) => H( P(A),c) }

Well, we ain't supposed to be doing homework problems here, but the proofs are extremely simple.

for proof 1: Your notation is inconsistent, so can't help until you clean that up.

for proof 2: $$\forall A (H(A,c) \Rightarrow H(P(A),c))$$
Assuming A, H(A,c), and P(A) exist immediately gives result H(P(A),c) from b 1 and 2.

 

[stauros]

Well, we ain't supposed to be doing homework problems here, but the proofs are extremely simple.

for proof 1: Your notation is inconsistent, so can't help until you clean that up.

for proof 2: $$\forall A (H(A,c) \Rightarrow H(P(A),c))$$
Assuming A, H(A,c), and P(A) exist immediately gives result H(P(A),c) from b 1 and 2.

Sorry to ask but do you know how a proof is done in predicate calculus?

I mean have you done any predicate calculus?

 

[xxxx0xxxx]

Sorry to ask but do you know how a proof is done in predicate calculus?

I mean have you done any predicate calculus?

Um yeah ...