Theorem soeq1 2141

Description: Equality theorem for the strict ordering predicate.

Assertion

Ref	Expression
soeq1	⊢ (R = S → (R Or A ↔ S Or A))

Proof of Theorem soeq1

Step	Hyp	Ref	Expression
1		poeq1 2130	. . 3 ⊢ (R = S → (R Po A ↔ S Po A))
2		breq 2064	. . . . . 6 ⊢ (R = S → (xRy ↔ xSy))
3		pm4.2i 149	. . . . . 6 ⊢ (R = S → (x = y ↔ x = y))
4		breq 2064	. . . . . 6 ⊢ (R = S → (yRx ↔ ySx))
5	2, 3, 4	bi3ord 635	. . . . 5 ⊢ (R = S → ((xRy ∨ x = y ∨ yRx) ↔ (xSy ∨ x = y ∨ ySx)))
6	5	biraldv 1219	. . . 4 ⊢ (R = S → (∀y ∈ A (xRy ∨ x = y ∨ yRx) ↔ ∀y ∈ A (xSy ∨ x = y ∨ ySx)))
7	6	biraldv 1219	. . 3 ⊢ (R = S → (∀x ∈ A ∀y ∈ A (xRy ∨ x = y ∨ yRx) ↔ ∀x ∈ A ∀y ∈ A (xSy ∨ x = y ∨ ySx)))
8	1, 7	anbi12d 476	. 2 ⊢ (R = S → ((R Po A ∧ ∀x ∈ A ∀y ∈ A (xRy ∨ x = y ∨ yRx)) ↔ (S Po A ∧ ∀x ∈ A ∀y ∈ A (xSy ∨ x = y ∨ ySx))))
9		df-so 2138	. 2 ⊢ (R Or A ↔ (R Po A ∧ ∀x ∈ A ∀y ∈ A (xRy ∨ x = y ∨ yRx)))
10		df-so 2138	. 2 ⊢ (S Or A ↔ (S Po A ∧ ∀x ∈ A ∀y ∈ A (xSy ∨ x = y ∨ ySx)))
11	8, 9, 10	3bitr4g 428	1 ⊢ (R = S → (R Or A ↔ S Or A))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196   ∨ w3o 580   = weq 797   = wceq 1091  ∀wral 1201   class class class wbr 2054   Po wpo 2058   Or wor 2059

This theorem is referenced by:  weeq1 2189

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-br 2063  df-po 2128  df-so 2138

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