Metamath Proof Explorer


Theorem dfssr2

Description: Alternate definition of the subset relation. (Contributed by Peter Mazsa, 9-Aug-2021)

Ref Expression
Assertion dfssr2 S = ( ( V × V ) ∖ ran ( E ⋉ ( V ∖ E ) ) )

Proof

Step Hyp Ref Expression
1 epel ( 𝑧 E 𝑥𝑧𝑥 )
2 brvdif ( 𝑧 ( V ∖ E ) 𝑦 ↔ ¬ 𝑧 E 𝑦 )
3 epel ( 𝑧 E 𝑦𝑧𝑦 )
4 2 3 xchbinx ( 𝑧 ( V ∖ E ) 𝑦 ↔ ¬ 𝑧𝑦 )
5 1 4 anbi12i ( ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) ↔ ( 𝑧𝑥 ∧ ¬ 𝑧𝑦 ) )
6 5 exbii ( ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) ↔ ∃ 𝑧 ( 𝑧𝑥 ∧ ¬ 𝑧𝑦 ) )
7 6 notbii ( ¬ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) ↔ ¬ ∃ 𝑧 ( 𝑧𝑥 ∧ ¬ 𝑧𝑦 ) )
8 dfss6 ( 𝑥𝑦 ↔ ¬ ∃ 𝑧 ( 𝑧𝑥 ∧ ¬ 𝑧𝑦 ) )
9 7 8 bitr4i ( ¬ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) ↔ 𝑥𝑦 )
10 9 opabbii { ⟨ 𝑥 , 𝑦 ⟩ ∣ ¬ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥𝑦 }
11 rnxrn ran ( E ⋉ ( V ∖ E ) ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) }
12 11 difeq2i ( ( V × V ) ∖ ran ( E ⋉ ( V ∖ E ) ) ) = ( ( V × V ) ∖ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) } )
13 vvdifopab ( ( V × V ) ∖ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) } ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ¬ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) }
14 12 13 eqtri ( ( V × V ) ∖ ran ( E ⋉ ( V ∖ E ) ) ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ¬ ∃ 𝑧 ( 𝑧 E 𝑥𝑧 ( V ∖ E ) 𝑦 ) }
15 df-ssr S = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥𝑦 }
16 10 14 15 3eqtr4ri S = ( ( V × V ) ∖ ran ( E ⋉ ( V ∖ E ) ) )