Metamath Proof Explorer


Theorem discsubclem

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

Ref Expression
Hypothesis discsubc.j 𝐽 = ( 𝑥𝑆 , 𝑦𝑆 ↦ if ( 𝑥 = 𝑦 , { ( 𝐼𝑥 ) } , ∅ ) )
Assertion discsubclem 𝐽 Fn ( 𝑆 × 𝑆 )

Proof

Step Hyp Ref Expression
1 discsubc.j 𝐽 = ( 𝑥𝑆 , 𝑦𝑆 ↦ if ( 𝑥 = 𝑦 , { ( 𝐼𝑥 ) } , ∅ ) )
2 snex { ( 𝐼𝑥 ) } ∈ V
3 0ex ∅ ∈ V
4 2 3 ifex if ( 𝑥 = 𝑦 , { ( 𝐼𝑥 ) } , ∅ ) ∈ V
5 1 4 fnmpoi 𝐽 Fn ( 𝑆 × 𝑆 )