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 x ↔ z ∈ x
2 brvdif ⊢ z V ∖ E y ↔ ¬ z E y
3 epel ⊢ z E y ↔ z ∈ y
4 2 3 xchbinx ⊢ z V ∖ E y ↔ ¬ z ∈ y
5 1 4 anbi12i ⊢ z E x ∧ z V ∖ E y ↔ z ∈ x ∧ ¬ z ∈ y
6 5 exbii ⊢ ∃ z z E x ∧ z V ∖ E y ↔ ∃ z z ∈ x ∧ ¬ z ∈ y
7 6 notbii ⊢ ¬ ∃ z z E x ∧ z V ∖ E y ↔ ¬ ∃ z z ∈ x ∧ ¬ z ∈ y
8 dfss6 ⊢ x ⊆ y ↔ ¬ ∃ z z ∈ x ∧ ¬ z ∈ y
9 7 8 bitr4i ⊢ ¬ ∃ z z E x ∧ z V ∖ E y ↔ x ⊆ y
10 9 opabbii ⊢ x y | ¬ ∃ z z E x ∧ z V ∖ E y = x y | x ⊆ y
11 rnxrn ⊢ ran ⁡ E ⋉ V ∖ E = x y | ∃ z z E x ∧ z V ∖ E y
12 11 difeq2i ⊢ V × V ∖ ran ⁡ E ⋉ V ∖ E = V × V ∖ x y | ∃ z z E x ∧ z V ∖ E y
13 vvdifopab ⊢ V × V ∖ x y | ∃ z z E x ∧ z V ∖ E y = x y | ¬ ∃ z z E x ∧ z V ∖ E y
14 12 13 eqtri ⊢ V × V ∖ ran ⁡ E ⋉ V ∖ E = x y | ¬ ∃ z z E x ∧ z V ∖ E y
15 df-ssr ⊢ S = x y | x ⊆ y
16 10 14 15 3eqtr4ri ⊢ S = V × V ∖ ran ⁡ E ⋉ V ∖ E