Metamath Proof Explorer


Theorem onsetrec

Description: Construct On using set recursion. When x e. On , the function F constructs the least ordinal greater than any of the elements of x , which is U. x for a limit ordinal and suc U. x for a successor ordinal.

For example, ( F{ 1o , 2o } ) = { U. { 1o , 2o } , suc U. { 1o , 2o } } = { 2o , 3o } which contains 3o , and ( F_om ) = { U.om , suc U. om } = {om , om +o 1o } , which contains _om . If we start with the empty set and keep applying F transfinitely many times, all ordinal numbers will be generated.

Any function F fulfilling lemmas onsetreclem2 and onsetreclem3 will recursively generate On ; for example, F = ( x e.V |-> suc suc U. x } ) also works. Whether this function or the function in the theorem is used, taking this theorem as a definition of On is unsatisfying because it relies on the different properties of limit and successor ordinals. A different approach could be to let F = ( x e. V |-> { y e. ~P x | Tr y } ) , based on dfon2 .

The proof of this theorem uses the dummy variable a rather than x to avoid a distinct variable condition between F and x . (Contributed by Emmett Weisz, 22-Jun-2021)

Ref Expression
Hypothesis onsetrec.1 F = x V x suc x
Assertion onsetrec setrecs F = On

Proof

Step Hyp Ref Expression
1 onsetrec.1 F = x V x suc x
2 eqid setrecs F = setrecs F
3 1 onsetreclem2 a On F a On
4 3 ax-gen a a On F a On
5 4 a1i a a On F a On
6 2 5 setrec2v setrecs F On
7 6 mptru setrecs F On
8 vex a V
9 8 a1i a setrecs F a V
10 id a setrecs F a setrecs F
11 2 9 10 setrec1 a setrecs F F a setrecs F
12 1 onsetreclem3 a On a F a
13 ssel F a setrecs F a F a a setrecs F
14 11 12 13 syl2im a setrecs F a On a setrecs F
15 14 com12 a On a setrecs F a setrecs F
16 15 rgen a On a setrecs F a setrecs F
17 tfi setrecs F On a On a setrecs F a setrecs F setrecs F = On
18 7 16 17 mp2an setrecs F = On