Metamath Proof Explorer


Theorem xporderlem

Description: Lemma for lexicographical ordering theorems. (Contributed by Scott Fenton, 16-Mar-2011)

Ref Expression
Hypothesis xporderlem.1 ⊢ T = x y | x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y
Assertion xporderlem ⊢ a b T c d ↔ a ∈ A ∧ c ∈ A ∧ b ∈ B ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d

Proof

Step Hyp Ref Expression
1 xporderlem.1 ⊢ T = x y | x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y
2 df-br ⊢ a b T c d ↔ a b c d ∈ T
3 1 eleq2i ⊢ a b c d ∈ T ↔ a b c d ∈ x y | x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y
4 2 3 bitri ⊢ a b T c d ↔ a b c d ∈ x y | x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y
5 opex ⊢ a b ∈ V
6 opex ⊢ c d ∈ V
7 eleq1 ⊢ x = a b → x ∈ A × B ↔ a b ∈ A × B
8 opelxp ⊢ a b ∈ A × B ↔ a ∈ A ∧ b ∈ B
9 7 8 bitrdi ⊢ x = a b → x ∈ A × B ↔ a ∈ A ∧ b ∈ B
10 9 anbi1d ⊢ x = a b → x ∈ A × B ∧ y ∈ A × B ↔ a ∈ A ∧ b ∈ B ∧ y ∈ A × B
11 vex ⊢ a ∈ V
12 vex ⊢ b ∈ V
13 11 12 op1std ⊢ x = a b → 1 st ⁡ x = a
14 13 breq1d ⊢ x = a b → 1 st ⁡ x R 1 st ⁡ y ↔ a R 1 st ⁡ y
15 13 eqeq1d ⊢ x = a b → 1 st ⁡ x = 1 st ⁡ y ↔ a = 1 st ⁡ y
16 11 12 op2ndd ⊢ x = a b → 2 nd ⁡ x = b
17 16 breq1d ⊢ x = a b → 2 nd ⁡ x S 2 nd ⁡ y ↔ b S 2 nd ⁡ y
18 15 17 anbi12d ⊢ x = a b → 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y ↔ a = 1 st ⁡ y ∧ b S 2 nd ⁡ y
19 14 18 orbi12d ⊢ x = a b → 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y ↔ a R 1 st ⁡ y ∨ a = 1 st ⁡ y ∧ b S 2 nd ⁡ y
20 10 19 anbi12d ⊢ x = a b → x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y ↔ a ∈ A ∧ b ∈ B ∧ y ∈ A × B ∧ a R 1 st ⁡ y ∨ a = 1 st ⁡ y ∧ b S 2 nd ⁡ y
21 eleq1 ⊢ y = c d → y ∈ A × B ↔ c d ∈ A × B
22 opelxp ⊢ c d ∈ A × B ↔ c ∈ A ∧ d ∈ B
23 21 22 bitrdi ⊢ y = c d → y ∈ A × B ↔ c ∈ A ∧ d ∈ B
24 23 anbi2d ⊢ y = c d → a ∈ A ∧ b ∈ B ∧ y ∈ A × B ↔ a ∈ A ∧ b ∈ B ∧ c ∈ A ∧ d ∈ B
25 vex ⊢ c ∈ V
26 vex ⊢ d ∈ V
27 25 26 op1std ⊢ y = c d → 1 st ⁡ y = c
28 27 breq2d ⊢ y = c d → a R 1 st ⁡ y ↔ a R c
29 27 eqeq2d ⊢ y = c d → a = 1 st ⁡ y ↔ a = c
30 25 26 op2ndd ⊢ y = c d → 2 nd ⁡ y = d
31 30 breq2d ⊢ y = c d → b S 2 nd ⁡ y ↔ b S d
32 29 31 anbi12d ⊢ y = c d → a = 1 st ⁡ y ∧ b S 2 nd ⁡ y ↔ a = c ∧ b S d
33 28 32 orbi12d ⊢ y = c d → a R 1 st ⁡ y ∨ a = 1 st ⁡ y ∧ b S 2 nd ⁡ y ↔ a R c ∨ a = c ∧ b S d
34 24 33 anbi12d ⊢ y = c d → a ∈ A ∧ b ∈ B ∧ y ∈ A × B ∧ a R 1 st ⁡ y ∨ a = 1 st ⁡ y ∧ b S 2 nd ⁡ y ↔ a ∈ A ∧ b ∈ B ∧ c ∈ A ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d
35 5 6 20 34 opelopab ⊢ a b c d ∈ x y | x ∈ A × B ∧ y ∈ A × B ∧ 1 st ⁡ x R 1 st ⁡ y ∨ 1 st ⁡ x = 1 st ⁡ y ∧ 2 nd ⁡ x S 2 nd ⁡ y ↔ a ∈ A ∧ b ∈ B ∧ c ∈ A ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d
36 an4 ⊢ a ∈ A ∧ b ∈ B ∧ c ∈ A ∧ d ∈ B ↔ a ∈ A ∧ c ∈ A ∧ b ∈ B ∧ d ∈ B
37 36 anbi1i ⊢ a ∈ A ∧ b ∈ B ∧ c ∈ A ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d ↔ a ∈ A ∧ c ∈ A ∧ b ∈ B ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d
38 4 35 37 3bitri ⊢ a b T c d ↔ a ∈ A ∧ c ∈ A ∧ b ∈ B ∧ d ∈ B ∧ a R c ∨ a = c ∧ b S d