Metamath Proof Explorer


Theorem discsubclem

Description: Lemma for discsubc . (Contributed by Zhi Wang, 1-Nov-2025)

Ref Expression
Hypothesis discsubc.j
|- J = ( x e. S , y e. S |-> if ( x = y , { ( I ` x ) } , (/) ) )
Assertion discsubclem
|- J Fn ( S X. S )

Proof

Step Hyp Ref Expression
1 discsubc.j
 |-  J = ( x e. S , y e. S |-> if ( x = y , { ( I ` x ) } , (/) ) )
2 snex
 |-  { ( I ` x ) } e. _V
3 0ex
 |-  (/) e. _V
4 2 3 ifex
 |-  if ( x = y , { ( I ` x ) } , (/) ) e. _V
5 1 4 fnmpoi
 |-  J Fn ( S X. S )