Formal Proof for Predicate Calculus 2: Solving Complex Operations and Predicates

Click For Summary
The discussion focuses on formal proof techniques in predicate calculus, specifically addressing complex operations and predicates. It outlines a set of assumptions involving one-place and two-place operations and predicates, leading to a formal proof requirement. Participants express frustration over the lack of progress in solving the proof, indicating that previous attempts were unsuccessful. There is a distinction made between the current problem and earlier discussions, emphasizing its complexity. The conversation highlights the challenges in deriving the necessary conclusions from the given assumptions.
solakis
Messages
19
Reaction score
0
Let:

1) P be one place operation

2) H be two place operation

3) G be two place predicate

4) k, m be two constants


Let :

The following assumptions :

1) \forall x [\neg G(x,k)\Longrightarrow G[H(P(x),x),m]]



2)\forall x\forall y\forall z[G(x,y)\Longrightarrow G[H(z,x),H(z,y)]]

3)\forall x\forall y\forall z [G(x,y)\wedge G(y,z)\Longrightarrow G(x,z)]

4)\forall x\forall y [G(x,y)\Longrightarrow G(y,x)]

5)\forall x\forall y [G[H(x,y),H(y,x)]]

6)\forall x[ G[H(x,m),m]]

Then formally prove that:

Then formally prove : \forall x\forall y\forall z[\neg G(x,k)\Longrightarrow(G[H(x,y),H(x,z)]\Longrightarrow G(y,z))]
 
Physics news on Phys.org
That's the same one as here, luckily you've formatted it a bit better this time (Y).

Any progress on the answer?
 
CompuChip said:
That's the same one as here, luckily you've formatted it a bit better this time (Y).

Any progress on the answer?

That is a completely different problem.

No, no any answer yet.
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
Replies
1
Views
1K