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 x ω f z x z f z z

Proof

Step Hyp Ref Expression
1 eqid x = x
2 eqid t ω , y x v t = t ω , y x v t
3 eqid w x u suc v -1 w = w x u suc v -1 w
4 1 2 3 axcclem x ω f z x z f z z