Metamath Proof Explorer


Theorem porpss

Description: Every class is partially ordered by proper subsets. (Contributed by Stefan O'Rear, 2-Nov-2014)

Ref Expression
Assertion porpss ⊢ [⊂] Po A

Proof

Step Hyp Ref Expression
1 pssirr ⊢ ¬ x ⊂ x
2 psstr ⊢ x ⊂ y ∧ y ⊂ z → x ⊂ z
3 vex ⊢ x ∈ V
4 3 brrpss ⊢ x [⊂] x ↔ x ⊂ x
5 4 notbii ⊢ ¬ x [⊂] x ↔ ¬ x ⊂ x
6 vex ⊢ y ∈ V
7 6 brrpss ⊢ x [⊂] y ↔ x ⊂ y
8 vex ⊢ z ∈ V
9 8 brrpss ⊢ y [⊂] z ↔ y ⊂ z
10 7 9 anbi12i ⊢ x [⊂] y ∧ y [⊂] z ↔ x ⊂ y ∧ y ⊂ z
11 8 brrpss ⊢ x [⊂] z ↔ x ⊂ z
12 10 11 imbi12i ⊢ x [⊂] y ∧ y [⊂] z → x [⊂] z ↔ x ⊂ y ∧ y ⊂ z → x ⊂ z
13 5 12 anbi12i ⊢ ¬ x [⊂] x ∧ x [⊂] y ∧ y [⊂] z → x [⊂] z ↔ ¬ x ⊂ x ∧ x ⊂ y ∧ y ⊂ z → x ⊂ z
14 1 2 13 mpbir2an ⊢ ¬ x [⊂] x ∧ x [⊂] y ∧ y [⊂] z → x [⊂] z
15 14 rgenw ⊢ ∀ z ∈ A ¬ x [⊂] x ∧ x [⊂] y ∧ y [⊂] z → x [⊂] z
16 15 rgen2w ⊢ ∀ x ∈ A ∀ y ∈ A ∀ z ∈ A ¬ x [⊂] x ∧ x [⊂] y ∧ y [⊂] z → x [⊂] z
17 df-po ⊢ [⊂] Po A ↔ ∀ x ∈ A ∀ y ∈ A ∀ z ∈ A ¬ x [⊂] x ∧ x [⊂] y ∧ y [⊂] z → x [⊂] z
18 16 17 mpbir ⊢ [⊂] Po A