Metamath Proof Explorer


Theorem heeq12

Description: Equality law for relations being herditary over a class. (Contributed by RP, 27-Mar-2020)

Ref Expression
Assertion heeq12 ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → ( 𝑅 hereditary 𝐴𝑆 hereditary 𝐵 ) )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → 𝑅 = 𝑆 )
2 simpr ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → 𝐴 = 𝐵 )
3 1 2 imaeq12d ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → ( 𝑅𝐴 ) = ( 𝑆𝐵 ) )
4 3 2 sseq12d ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → ( ( 𝑅𝐴 ) ⊆ 𝐴 ↔ ( 𝑆𝐵 ) ⊆ 𝐵 ) )
5 df-he ( 𝑅 hereditary 𝐴 ↔ ( 𝑅𝐴 ) ⊆ 𝐴 )
6 df-he ( 𝑆 hereditary 𝐵 ↔ ( 𝑆𝐵 ) ⊆ 𝐵 )
7 4 5 6 3bitr4g ( ( 𝑅 = 𝑆𝐴 = 𝐵 ) → ( 𝑅 hereditary 𝐴𝑆 hereditary 𝐵 ) )