Metamath Proof Explorer


Theorem axcontlem5

Description: Lemma for axcont . Compute the value of F . (Contributed by Scott Fenton, 18-Jun-2013)

Ref Expression
Hypotheses axcontlem5.1 ⊢ D = p ∈ 𝔼 ⁡ N | U Btwn Z p ∨ p Btwn Z U
axcontlem5.2 ⊢ F = x t | x ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
Assertion axcontlem5 ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → F ⁡ P = T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i

Proof

Step Hyp Ref Expression
1 axcontlem5.1 ⊢ D = p ∈ 𝔼 ⁡ N | U Btwn Z p ∨ p Btwn Z U
2 axcontlem5.2 ⊢ F = x t | x ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
3 1 2 axcontlem2 ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U → F : D ⟶ 1-1 onto 0 +∞
4 f1of ⊢ F : D ⟶ 1-1 onto 0 +∞ → F : D ⟶ 0 +∞
5 3 4 syl ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U → F : D ⟶ 0 +∞
6 5 ffvelcdmda ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → F ⁡ P ∈ 0 +∞
7 eleq1 ⊢ F ⁡ P = T → F ⁡ P ∈ 0 +∞ ↔ T ∈ 0 +∞
8 6 7 syl5ibcom ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → F ⁡ P = T → T ∈ 0 +∞
9 simpl ⊢ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i → T ∈ 0 +∞
10 9 a1i ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i → T ∈ 0 +∞
11 f1ofn ⊢ F : D ⟶ 1-1 onto 0 +∞ → F Fn D
12 3 11 syl ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U → F Fn D
13 fnbrfvb ⊢ F Fn D ∧ P ∈ D → F ⁡ P = T ↔ P F T
14 12 13 sylan ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → F ⁡ P = T ↔ P F T
15 14 3adant3 ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D ∧ T ∈ 0 +∞ → F ⁡ P = T ↔ P F T
16 eleq1 ⊢ x = P → x ∈ D ↔ P ∈ D
17 fveq1 ⊢ x = P → x ⁡ i = P ⁡ i
18 17 eqeq1d ⊢ x = P → x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
19 18 ralbidv ⊢ x = P → ∀ i ∈ 1 … N x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
20 19 anbi2d ⊢ x = P → t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
21 16 20 anbi12d ⊢ x = P → x ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N x ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ P ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i
22 eleq1 ⊢ t = T → t ∈ 0 +∞ ↔ T ∈ 0 +∞
23 oveq2 ⊢ t = T → 1 − t = 1 − T
24 23 oveq1d ⊢ t = T → 1 − t ⁢ Z ⁡ i = 1 − T ⁢ Z ⁡ i
25 oveq1 ⊢ t = T → t ⁢ U ⁡ i = T ⁢ U ⁡ i
26 24 25 oveq12d ⊢ t = T → 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
27 26 eqeq2d ⊢ t = T → P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
28 27 ralbidv ⊢ t = T → ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
29 22 28 anbi12d ⊢ t = T → t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
30 29 anbi2d ⊢ t = T → P ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
31 anass ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞
32 anidm ⊢ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ↔ T ∈ 0 +∞
33 32 anbi2i ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ↔ P ∈ D ∧ T ∈ 0 +∞
34 31 33 bitr2i ⊢ P ∈ D ∧ T ∈ 0 +∞ ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞
35 34 anbi1i ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
36 anass ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
37 anass ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
38 35 36 37 3bitr3i ⊢ P ∈ D ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
39 30 38 bitrdi ⊢ t = T → P ∈ D ∧ t ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − t ⁢ Z ⁡ i + t ⁢ U ⁡ i ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
40 21 39 2 brabg ⊢ P ∈ D ∧ T ∈ 0 +∞ → P F T ↔ P ∈ D ∧ T ∈ 0 +∞ ∧ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
41 40 bianabs ⊢ P ∈ D ∧ T ∈ 0 +∞ → P F T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
42 41 3adant1 ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D ∧ T ∈ 0 +∞ → P F T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
43 15 42 bitrd ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D ∧ T ∈ 0 +∞ → F ⁡ P = T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
44 43 3expia ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → T ∈ 0 +∞ → F ⁡ P = T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i
45 8 10 44 pm5.21ndd ⊢ N ∈ ℕ ∧ Z ∈ 𝔼 ⁡ N ∧ U ∈ 𝔼 ⁡ N ∧ Z ≠ U ∧ P ∈ D → F ⁡ P = T ↔ T ∈ 0 +∞ ∧ ∀ i ∈ 1 … N P ⁡ i = 1 − T ⁢ Z ⁡ i + T ⁢ U ⁡ i