Metamath Proof Explorer


Theorem ndmaovrcl

Description: Reverse closure law, in contrast to ndmovrcl where it is required that the operation's domain doesn't contain the empty set ( -. (/) e. S ), no additional asumption is required. (Contributed by Alexander van der Vekens, 26-May-2017)

Ref Expression
Hypothesis ndmaov.1 dom 𝐹 = ( 𝑆 × 𝑆 )
Assertion ndmaovrcl ( (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 → ( 𝐴𝑆𝐵𝑆 ) )

Proof

Step Hyp Ref Expression
1 ndmaov.1 dom 𝐹 = ( 𝑆 × 𝑆 )
2 aovvdm ( (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 → ⟨ 𝐴 , 𝐵 ⟩ ∈ dom 𝐹 )
3 opelxp ( ⟨ 𝐴 , 𝐵 ⟩ ∈ ( 𝑆 × 𝑆 ) ↔ ( 𝐴𝑆𝐵𝑆 ) )
4 3 biimpi ( ⟨ 𝐴 , 𝐵 ⟩ ∈ ( 𝑆 × 𝑆 ) → ( 𝐴𝑆𝐵𝑆 ) )
5 4 1 eleq2s ( ⟨ 𝐴 , 𝐵 ⟩ ∈ dom 𝐹 → ( 𝐴𝑆𝐵𝑆 ) )
6 2 5 syl ( (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 → ( 𝐴𝑆𝐵𝑆 ) )