Metamath Proof Explorer


Theorem discsubclem

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

Ref Expression
Hypothesis discsubc.j J = x S , y S if x = y I x
Assertion discsubclem J Fn S × S

Proof

Step Hyp Ref Expression
1 discsubc.j J = x S , y S if x = y I x
2 snex I x V
3 0ex V
4 2 3 ifex if x = y I x V
5 1 4 fnmpoi J Fn S × S