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 ...