Metamath Proof Explorer


Theorem ssxr

Description: The three (non-exclusive) possibilities implied by a subset of extended reals. (Contributed by NM, 25-Oct-2005) (Proof shortened by Andrew Salmon, 19-Nov-2011)

Ref Expression
Assertion ssxr ⊢ A ⊆ ℝ * → A ⊆ ℝ ∨ +∞ ∈ A ∨ −∞ ∈ A

Proof

Step Hyp Ref Expression
1 df-pr ⊢ +∞ −∞ = +∞ ∪ −∞
2 1 ineq2i ⊢ A ∩ +∞ −∞ = A ∩ +∞ ∪ −∞
3 indi ⊢ A ∩ +∞ ∪ −∞ = A ∩ +∞ ∪ A ∩ −∞
4 2 3 eqtri ⊢ A ∩ +∞ −∞ = A ∩ +∞ ∪ A ∩ −∞
5 disjsn ⊢ A ∩ +∞ = ∅ ↔ ¬ +∞ ∈ A
6 disjsn ⊢ A ∩ −∞ = ∅ ↔ ¬ −∞ ∈ A
7 5 6 anbi12i ⊢ A ∩ +∞ = ∅ ∧ A ∩ −∞ = ∅ ↔ ¬ +∞ ∈ A ∧ ¬ −∞ ∈ A
8 7 biimpri ⊢ ¬ +∞ ∈ A ∧ ¬ −∞ ∈ A → A ∩ +∞ = ∅ ∧ A ∩ −∞ = ∅
9 pm4.56 ⊢ ¬ +∞ ∈ A ∧ ¬ −∞ ∈ A ↔ ¬ +∞ ∈ A ∨ −∞ ∈ A
10 un00 ⊢ A ∩ +∞ = ∅ ∧ A ∩ −∞ = ∅ ↔ A ∩ +∞ ∪ A ∩ −∞ = ∅
11 8 9 10 3imtr3i ⊢ ¬ +∞ ∈ A ∨ −∞ ∈ A → A ∩ +∞ ∪ A ∩ −∞ = ∅
12 4 11 eqtrid ⊢ ¬ +∞ ∈ A ∨ −∞ ∈ A → A ∩ +∞ −∞ = ∅
13 reldisj ⊢ A ⊆ ℝ ∪ +∞ −∞ → A ∩ +∞ −∞ = ∅ ↔ A ⊆ ℝ ∪ +∞ −∞ ∖ +∞ −∞
14 renfdisj ⊢ ℝ ∩ +∞ −∞ = ∅
15 disj3 ⊢ ℝ ∩ +∞ −∞ = ∅ ↔ ℝ = ℝ ∖ +∞ −∞
16 14 15 mpbi ⊢ ℝ = ℝ ∖ +∞ −∞
17 difun2 ⊢ ℝ ∪ +∞ −∞ ∖ +∞ −∞ = ℝ ∖ +∞ −∞
18 16 17 eqtr4i ⊢ ℝ = ℝ ∪ +∞ −∞ ∖ +∞ −∞
19 18 sseq2i ⊢ A ⊆ ℝ ↔ A ⊆ ℝ ∪ +∞ −∞ ∖ +∞ −∞
20 13 19 bitr4di ⊢ A ⊆ ℝ ∪ +∞ −∞ → A ∩ +∞ −∞ = ∅ ↔ A ⊆ ℝ
21 12 20 imbitrid ⊢ A ⊆ ℝ ∪ +∞ −∞ → ¬ +∞ ∈ A ∨ −∞ ∈ A → A ⊆ ℝ
22 21 orrd ⊢ A ⊆ ℝ ∪ +∞ −∞ → +∞ ∈ A ∨ −∞ ∈ A ∨ A ⊆ ℝ
23 df-xr ⊢ ℝ * = ℝ ∪ +∞ −∞
24 23 sseq2i ⊢ A ⊆ ℝ * ↔ A ⊆ ℝ ∪ +∞ −∞
25 3orrot ⊢ A ⊆ ℝ ∨ +∞ ∈ A ∨ −∞ ∈ A ↔ +∞ ∈ A ∨ −∞ ∈ A ∨ A ⊆ ℝ
26 df-3or ⊢ +∞ ∈ A ∨ −∞ ∈ A ∨ A ⊆ ℝ ↔ +∞ ∈ A ∨ −∞ ∈ A ∨ A ⊆ ℝ
27 25 26 bitri ⊢ A ⊆ ℝ ∨ +∞ ∈ A ∨ −∞ ∈ A ↔ +∞ ∈ A ∨ −∞ ∈ A ∨ A ⊆ ℝ
28 22 24 27 3imtr4i ⊢ A ⊆ ℝ * → A ⊆ ℝ ∨ +∞ ∈ A ∨ −∞ ∈ A