[General boards] [Fall 2018 courses] [Summer 2018 courses] [Winter 2018 courses] [Older or newer terms]

Confusion about One-Point Law


#1

I’m confused about this problem from the textbook. While I understand that One-Point Law is used to substitute values, there is no relation between the x prime and the y prime. How was the One-Point Law applied twice here?

Any help will be appreciated


#2

The top line is
∀x, y· ∃x’, y’· x’=x+1 ∧ y’=y
The one-point law I am using is ∃a· a=b ∧ P == (P but replace a with b)
So let’s use it on ∃y’· x’=x+1 ∧ y’=y
We get (x’=x+1 but replace y’ with y)
Replacement is easy because y’ does not occur in x’=x+1 so replacement leaves it the same. We get
∃x’· x’=x+1
which is the same as
∃x’· x’=x+1 ∧ ⊤
Now use one-point again and get
(⊤ but replace x’ with x+1)
and that’s easy because x’ does not occur in ⊤. We get ⊤.