Metamath Proof Explorer


Theorem sucneqoni

Description: Inequality of an ordinal set with its successor. Does not use the axiom of regularity. (Contributed by ML, 18-Oct-2020)

Ref Expression
Hypotheses sucneqoni.1
|- X = suc Y
sucneqoni.2
|- Y e. On
Assertion sucneqoni
|- X =/= Y

Proof

Step Hyp Ref Expression
1 sucneqoni.1
 |-  X = suc Y
2 sucneqoni.2
 |-  Y e. On
3 1 a1i
 |-  ( T. -> X = suc Y )
4 2 a1i
 |-  ( T. -> Y e. On )
5 3 4 sucneqond
 |-  ( T. -> X =/= Y )
6 5 mptru
 |-  X =/= Y