Metamath Proof Explorer


Theorem xrsupssd

Description: Inequality deduction for supremum of an extended real subset. (Contributed by Thierry Arnoux, 21-Mar-2017)

Ref Expression
Hypotheses xrsupssd.1 ⊢ φ → B ⊆ C
xrsupssd.2 ⊢ φ → C ⊆ ℝ *
Assertion xrsupssd ⊢ φ → sup B ℝ * < ≤ sup C ℝ * <

Proof

Step Hyp Ref Expression
1 xrsupssd.1 ⊢ φ → B ⊆ C
2 xrsupssd.2 ⊢ φ → C ⊆ ℝ *
3 xrltso ⊢ < Or ℝ *
4 3 a1i ⊢ φ → < Or ℝ *
5 1 2 sstrd ⊢ φ → B ⊆ ℝ *
6 xrsupss ⊢ B ⊆ ℝ * → ∃ x ∈ ℝ * ∀ y ∈ B ¬ x < y ∧ ∀ y ∈ ℝ * y < x → ∃ z ∈ B y < z
7 5 6 syl ⊢ φ → ∃ x ∈ ℝ * ∀ y ∈ B ¬ x < y ∧ ∀ y ∈ ℝ * y < x → ∃ z ∈ B y < z
8 xrsupss ⊢ C ⊆ ℝ * → ∃ x ∈ ℝ * ∀ y ∈ C ¬ x < y ∧ ∀ y ∈ ℝ * y < x → ∃ z ∈ C y < z
9 2 8 syl ⊢ φ → ∃ x ∈ ℝ * ∀ y ∈ C ¬ x < y ∧ ∀ y ∈ ℝ * y < x → ∃ z ∈ C y < z
10 4 1 2 7 9 supssd ⊢ φ → ¬ sup C ℝ * < < sup B ℝ * <
11 4 7 supcl ⊢ φ → sup B ℝ * < ∈ ℝ *
12 4 9 supcl ⊢ φ → sup C ℝ * < ∈ ℝ *
13 xrlenlt ⊢ sup B ℝ * < ∈ ℝ * ∧ sup C ℝ * < ∈ ℝ * → sup B ℝ * < ≤ sup C ℝ * < ↔ ¬ sup C ℝ * < < sup B ℝ * <
14 11 12 13 syl2anc ⊢ φ → sup B ℝ * < ≤ sup C ℝ * < ↔ ¬ sup C ℝ * < < sup B ℝ * <
15 10 14 mpbird ⊢ φ → sup B ℝ * < ≤ sup C ℝ * <