Theorem poeq1 2130

Description: Equality theorem for partial ordering predicate.

Assertion

Ref	Expression
poeq1	⊢ (R = S → (R Po A ↔ S Po A))

Proof of Theorem poeq1

Step	Hyp	Ref	Expression
1		breq 2064	. . . . . . 7 ⊢ (R = S → (xRx ↔ xSx))
2	1	negbid 463	. . . . . 6 ⊢ (R = S → (¬ xRx ↔ ¬ xSx))
3		breq 2064	. . . . . . . 8 ⊢ (R = S → (xRy ↔ xSy))
4		breq 2064	. . . . . . . 8 ⊢ (R = S → (yRz ↔ ySz))
5	3, 4	anbi12d 476	. . . . . . 7 ⊢ (R = S → ((xRy ∧ yRz) ↔ (xSy ∧ ySz)))
6		breq 2064	. . . . . . 7 ⊢ (R = S → (xRz ↔ xSz))
7	5, 6	imbi12d 474	. . . . . 6 ⊢ (R = S → (((xRy ∧ yRz) → xRz) ↔ ((xSy ∧ ySz) → xSz)))
8	2, 7	anbi12d 476	. . . . 5 ⊢ (R = S → ((¬ xRx ∧ ((xRy ∧ yRz) → xRz)) ↔ (¬ xSx ∧ ((xSy ∧ ySz) → xSz))))
9	8	biraldv 1219	. . . 4 ⊢ (R = S → (∀z ∈ A (¬ xRx ∧ ((xRy ∧ yRz) → xRz)) ↔ ∀z ∈ A (¬ xSx ∧ ((xSy ∧ ySz) → xSz))))
10	9	biraldv 1219	. . 3 ⊢ (R = S → (∀y ∈ A ∀z ∈ A (¬ xRx ∧ ((xRy ∧ yRz) → xRz)) ↔ ∀y ∈ A ∀z ∈ A (¬ xSx ∧ ((xSy ∧ ySz) → xSz))))
11	10	biraldv 1219	. 2 ⊢ (R = S → (∀x ∈ A ∀y ∈ A ∀z ∈ A (¬ xRx ∧ ((xRy ∧ yRz) → xRz)) ↔ ∀x ∈ A ∀y ∈ A ∀z ∈ A (¬ xSx ∧ ((xSy ∧ ySz) → xSz))))
12		df-po 2128	. 2 ⊢ (R Po A ↔ ∀x ∈ A ∀y ∈ A ∀z ∈ A (¬ xRx ∧ ((xRy ∧ yRz) → xRz)))
13		df-po 2128	. 2 ⊢ (S Po A ↔ ∀x ∈ A ∀y ∈ A ∀z ∈ A (¬ xSx ∧ ((xSy ∧ ySz) → xSz)))
14	11, 12, 13	3bitr4g 428	1 ⊢ (R = S → (R Po A ↔ S Po A))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196   = wceq 1091  ∀wral 1201   class class class wbr 2054   Po wpo 2058

This theorem is referenced by:  soeq1 2141

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-an 198  df-ex 679  df-cleq 1097  df-clel 1099  df-ral 1205  df-br 2063  df-po 2128

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