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Metamath Proof Explorer

Theorem refreleq

Description: Equality theorem for reflexive relation. (Contributed by Peter Mazsa, 15-Apr-2019) (Revised by Peter Mazsa, 23-Sep-2021)

Ref Expression
Assertion refreleq ( 𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆 ) )

Proof

Step Hyp Ref Expression
1 dmeq ⊢ ( 𝑅 = 𝑆 → dom 𝑅 = dom 𝑆 )
2 rneq ⊢ ( 𝑅 = 𝑆 → ran 𝑅 = ran 𝑆 )
3 1 2 xpeq12d ⊢ ( 𝑅 = 𝑆 → ( dom 𝑅 × ran 𝑅 ) = ( dom 𝑆 × ran 𝑆 ) )
4 3 ineq2d ⊢ ( 𝑅 = 𝑆 → ( I ∩ ( dom 𝑅 × ran 𝑅 ) ) = ( I ∩ ( dom 𝑆 × ran 𝑆 ) ) )
5 id ⊢ ( 𝑅 = 𝑆 → 𝑅 = 𝑆 )
6 4 5 sseq12d ⊢ ( 𝑅 = 𝑆 → ( ( I ∩ ( dom 𝑅 × ran 𝑅 ) ) ⊆ 𝑅 ↔ ( I ∩ ( dom 𝑆 × ran 𝑆 ) ) ⊆ 𝑆 ) )
7 releq ⊢ ( 𝑅 = 𝑆 → ( Rel 𝑅 ↔ Rel 𝑆 ) )
8 6 7 anbi12d ⊢ ( 𝑅 = 𝑆 → ( ( ( I ∩ ( dom 𝑅 × ran 𝑅 ) ) ⊆ 𝑅 ∧ Rel 𝑅 ) ↔ ( ( I ∩ ( dom 𝑆 × ran 𝑆 ) ) ⊆ 𝑆 ∧ Rel 𝑆 ) ) )
9 dfrefrel2 ⊢ ( RefRel 𝑅 ↔ ( ( I ∩ ( dom 𝑅 × ran 𝑅 ) ) ⊆ 𝑅 ∧ Rel 𝑅 ) )
10 dfrefrel2 ⊢ ( RefRel 𝑆 ↔ ( ( I ∩ ( dom 𝑆 × ran 𝑆 ) ) ⊆ 𝑆 ∧ Rel 𝑆 ) )
11 8 9 10 3bitr4g ⊢ ( 𝑅 = 𝑆 → ( RefRel 𝑅 ↔ RefRel 𝑆 ) )