Metamath Proof Explorer


Theorem wemaplem1

Description: Value of the lexicographic order on a sequence space. (Contributed by Stefan O'Rear, 18-Jan-2015)

Ref Expression
Hypothesis wemapso.t ⊢ T = x y | ∃ z ∈ A x ⁡ z S y ⁡ z ∧ ∀ w ∈ A w R z → x ⁡ w = y ⁡ w
Assertion wemaplem1 ⊢ P ∈ V ∧ Q ∈ W → P T Q ↔ ∃ a ∈ A P ⁡ a S Q ⁡ a ∧ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b

Proof

Step Hyp Ref Expression
1 wemapso.t ⊢ T = x y | ∃ z ∈ A x ⁡ z S y ⁡ z ∧ ∀ w ∈ A w R z → x ⁡ w = y ⁡ w
2 fveq1 ⊢ x = P → x ⁡ z = P ⁡ z
3 fveq1 ⊢ y = Q → y ⁡ z = Q ⁡ z
4 2 3 breqan12d ⊢ x = P ∧ y = Q → x ⁡ z S y ⁡ z ↔ P ⁡ z S Q ⁡ z
5 fveq1 ⊢ x = P → x ⁡ w = P ⁡ w
6 fveq1 ⊢ y = Q → y ⁡ w = Q ⁡ w
7 5 6 eqeqan12d ⊢ x = P ∧ y = Q → x ⁡ w = y ⁡ w ↔ P ⁡ w = Q ⁡ w
8 7 imbi2d ⊢ x = P ∧ y = Q → w R z → x ⁡ w = y ⁡ w ↔ w R z → P ⁡ w = Q ⁡ w
9 8 ralbidv ⊢ x = P ∧ y = Q → ∀ w ∈ A w R z → x ⁡ w = y ⁡ w ↔ ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w
10 4 9 anbi12d ⊢ x = P ∧ y = Q → x ⁡ z S y ⁡ z ∧ ∀ w ∈ A w R z → x ⁡ w = y ⁡ w ↔ P ⁡ z S Q ⁡ z ∧ ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w
11 10 rexbidv ⊢ x = P ∧ y = Q → ∃ z ∈ A x ⁡ z S y ⁡ z ∧ ∀ w ∈ A w R z → x ⁡ w = y ⁡ w ↔ ∃ z ∈ A P ⁡ z S Q ⁡ z ∧ ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w
12 fveq2 ⊢ z = a → P ⁡ z = P ⁡ a
13 fveq2 ⊢ z = a → Q ⁡ z = Q ⁡ a
14 12 13 breq12d ⊢ z = a → P ⁡ z S Q ⁡ z ↔ P ⁡ a S Q ⁡ a
15 breq2 ⊢ z = a → w R z ↔ w R a
16 15 imbi1d ⊢ z = a → w R z → P ⁡ w = Q ⁡ w ↔ w R a → P ⁡ w = Q ⁡ w
17 16 ralbidv ⊢ z = a → ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w ↔ ∀ w ∈ A w R a → P ⁡ w = Q ⁡ w
18 breq1 ⊢ w = b → w R a ↔ b R a
19 fveq2 ⊢ w = b → P ⁡ w = P ⁡ b
20 fveq2 ⊢ w = b → Q ⁡ w = Q ⁡ b
21 19 20 eqeq12d ⊢ w = b → P ⁡ w = Q ⁡ w ↔ P ⁡ b = Q ⁡ b
22 18 21 imbi12d ⊢ w = b → w R a → P ⁡ w = Q ⁡ w ↔ b R a → P ⁡ b = Q ⁡ b
23 22 cbvralvw ⊢ ∀ w ∈ A w R a → P ⁡ w = Q ⁡ w ↔ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b
24 17 23 bitrdi ⊢ z = a → ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w ↔ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b
25 14 24 anbi12d ⊢ z = a → P ⁡ z S Q ⁡ z ∧ ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w ↔ P ⁡ a S Q ⁡ a ∧ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b
26 25 cbvrexvw ⊢ ∃ z ∈ A P ⁡ z S Q ⁡ z ∧ ∀ w ∈ A w R z → P ⁡ w = Q ⁡ w ↔ ∃ a ∈ A P ⁡ a S Q ⁡ a ∧ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b
27 11 26 bitrdi ⊢ x = P ∧ y = Q → ∃ z ∈ A x ⁡ z S y ⁡ z ∧ ∀ w ∈ A w R z → x ⁡ w = y ⁡ w ↔ ∃ a ∈ A P ⁡ a S Q ⁡ a ∧ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b
28 27 1 brabga ⊢ P ∈ V ∧ Q ∈ W → P T Q ↔ ∃ a ∈ A P ⁡ a S Q ⁡ a ∧ ∀ b ∈ A b R a → P ⁡ b = Q ⁡ b