Metamath Proof Explorer


Theorem supmullem1

Description: Lemma for supmul . (Contributed by Mario Carneiro, 5-Jul-2013)

Ref Expression
Hypotheses supmul.1 ⊢ C = z | ∃ v ∈ A ∃ b ∈ B z = v ⁢ b
supmul.2 ⊢ φ ↔ ∀ x ∈ A 0 ≤ x ∧ ∀ x ∈ B 0 ≤ x ∧ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x ∧ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x
Assertion supmullem1 ⊢ φ → ∀ w ∈ C w ≤ sup A ℝ < ⁢ sup B ℝ <

Proof

Step Hyp Ref Expression
1 supmul.1 ⊢ C = z | ∃ v ∈ A ∃ b ∈ B z = v ⁢ b
2 supmul.2 ⊢ φ ↔ ∀ x ∈ A 0 ≤ x ∧ ∀ x ∈ B 0 ≤ x ∧ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x ∧ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x
3 vex ⊢ w ∈ V
4 oveq1 ⊢ v = a → v ⁢ b = a ⁢ b
5 4 eqeq2d ⊢ v = a → z = v ⁢ b ↔ z = a ⁢ b
6 5 rexbidv ⊢ v = a → ∃ b ∈ B z = v ⁢ b ↔ ∃ b ∈ B z = a ⁢ b
7 6 cbvrexvw ⊢ ∃ v ∈ A ∃ b ∈ B z = v ⁢ b ↔ ∃ a ∈ A ∃ b ∈ B z = a ⁢ b
8 eqeq1 ⊢ z = w → z = a ⁢ b ↔ w = a ⁢ b
9 8 2rexbidv ⊢ z = w → ∃ a ∈ A ∃ b ∈ B z = a ⁢ b ↔ ∃ a ∈ A ∃ b ∈ B w = a ⁢ b
10 7 9 bitrid ⊢ z = w → ∃ v ∈ A ∃ b ∈ B z = v ⁢ b ↔ ∃ a ∈ A ∃ b ∈ B w = a ⁢ b
11 3 10 1 elab2 ⊢ w ∈ C ↔ ∃ a ∈ A ∃ b ∈ B w = a ⁢ b
12 2 simp2bi ⊢ φ → A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x
13 12 simp1d ⊢ φ → A ⊆ ℝ
14 13 sselda ⊢ φ ∧ a ∈ A → a ∈ ℝ
15 14 adantrr ⊢ φ ∧ a ∈ A ∧ b ∈ B → a ∈ ℝ
16 suprcl ⊢ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x → sup A ℝ < ∈ ℝ
17 12 16 syl ⊢ φ → sup A ℝ < ∈ ℝ
18 17 adantr ⊢ φ ∧ a ∈ A ∧ b ∈ B → sup A ℝ < ∈ ℝ
19 2 simp3bi ⊢ φ → B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x
20 19 simp1d ⊢ φ → B ⊆ ℝ
21 20 sselda ⊢ φ ∧ b ∈ B → b ∈ ℝ
22 21 adantrl ⊢ φ ∧ a ∈ A ∧ b ∈ B → b ∈ ℝ
23 suprcl ⊢ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x → sup B ℝ < ∈ ℝ
24 19 23 syl ⊢ φ → sup B ℝ < ∈ ℝ
25 24 adantr ⊢ φ ∧ a ∈ A ∧ b ∈ B → sup B ℝ < ∈ ℝ
26 simp1l ⊢ ∀ x ∈ A 0 ≤ x ∧ ∀ x ∈ B 0 ≤ x ∧ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x ∧ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x → ∀ x ∈ A 0 ≤ x
27 2 26 sylbi ⊢ φ → ∀ x ∈ A 0 ≤ x
28 breq2 ⊢ x = a → 0 ≤ x ↔ 0 ≤ a
29 28 rspccv ⊢ ∀ x ∈ A 0 ≤ x → a ∈ A → 0 ≤ a
30 27 29 syl ⊢ φ → a ∈ A → 0 ≤ a
31 30 imp ⊢ φ ∧ a ∈ A → 0 ≤ a
32 31 adantrr ⊢ φ ∧ a ∈ A ∧ b ∈ B → 0 ≤ a
33 simp1r ⊢ ∀ x ∈ A 0 ≤ x ∧ ∀ x ∈ B 0 ≤ x ∧ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x ∧ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x → ∀ x ∈ B 0 ≤ x
34 2 33 sylbi ⊢ φ → ∀ x ∈ B 0 ≤ x
35 breq2 ⊢ x = b → 0 ≤ x ↔ 0 ≤ b
36 35 rspccv ⊢ ∀ x ∈ B 0 ≤ x → b ∈ B → 0 ≤ b
37 34 36 syl ⊢ φ → b ∈ B → 0 ≤ b
38 37 imp ⊢ φ ∧ b ∈ B → 0 ≤ b
39 38 adantrl ⊢ φ ∧ a ∈ A ∧ b ∈ B → 0 ≤ b
40 suprub ⊢ A ⊆ ℝ ∧ A ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ A y ≤ x ∧ a ∈ A → a ≤ sup A ℝ <
41 12 40 sylan ⊢ φ ∧ a ∈ A → a ≤ sup A ℝ <
42 41 adantrr ⊢ φ ∧ a ∈ A ∧ b ∈ B → a ≤ sup A ℝ <
43 suprub ⊢ B ⊆ ℝ ∧ B ≠ ∅ ∧ ∃ x ∈ ℝ ∀ y ∈ B y ≤ x ∧ b ∈ B → b ≤ sup B ℝ <
44 19 43 sylan ⊢ φ ∧ b ∈ B → b ≤ sup B ℝ <
45 44 adantrl ⊢ φ ∧ a ∈ A ∧ b ∈ B → b ≤ sup B ℝ <
46 15 18 22 25 32 39 42 45 lemul12ad ⊢ φ ∧ a ∈ A ∧ b ∈ B → a ⁢ b ≤ sup A ℝ < ⁢ sup B ℝ <
47 46 ex ⊢ φ → a ∈ A ∧ b ∈ B → a ⁢ b ≤ sup A ℝ < ⁢ sup B ℝ <
48 breq1 ⊢ w = a ⁢ b → w ≤ sup A ℝ < ⁢ sup B ℝ < ↔ a ⁢ b ≤ sup A ℝ < ⁢ sup B ℝ <
49 48 biimprcd ⊢ a ⁢ b ≤ sup A ℝ < ⁢ sup B ℝ < → w = a ⁢ b → w ≤ sup A ℝ < ⁢ sup B ℝ <
50 47 49 syl6 ⊢ φ → a ∈ A ∧ b ∈ B → w = a ⁢ b → w ≤ sup A ℝ < ⁢ sup B ℝ <
51 50 rexlimdvv ⊢ φ → ∃ a ∈ A ∃ b ∈ B w = a ⁢ b → w ≤ sup A ℝ < ⁢ sup B ℝ <
52 11 51 biimtrid ⊢ φ → w ∈ C → w ≤ sup A ℝ < ⁢ sup B ℝ <
53 52 ralrimiv ⊢ φ → ∀ w ∈ C w ≤ sup A ℝ < ⁢ sup B ℝ <