Metamath Proof Explorer


Theorem axcc

Description: Although CC can be proven trivially using ac5 , we prove it here using DC. (New usage is discouraged.) (Contributed by Mario Carneiro, 2-Feb-2013)

Ref Expression
Assertion axcc xωfzxzfzz

Proof

Step Hyp Ref Expression
1 eqid x=x
2 eqid tω,yxvt=tω,yxvt
3 eqid wxusucv-1w=wxusucv-1w
4 1 2 3 axcclem xωfzxzfzz