Theorem poeq1 4334

Description: Equality theorem for partial ordering predicate. (Contributed by NM, 27-Mar-1997.)

Assertion

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

Proof of Theorem poeq1

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		breq 4035	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑥 ↔ 𝑥𝑆𝑥))
2	1	notbid 668	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑥 ↔ ¬ 𝑥𝑆𝑥))
3		breq 4035	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
4		breq 4035	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑧 ↔ 𝑦𝑆𝑧))
5	3, 4	anbi12d 473	. . . . . 6 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧)))
6		breq 4035	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
7	5, 6	imbi12d 234	. . . . 5 ⊢ (𝑅 = 𝑆 → (((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
8	2, 7	anbi12d 473	. . . 4 ⊢ (𝑅 = 𝑆 → ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
9	8	ralbidv 2497	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
10	9	2ralbidv 2521	. 2 ⊢ (𝑅 = 𝑆 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
11		df-po 4331	. 2 ⊢ (𝑅 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12		df-po 4331	. 2 ⊢ (𝑆 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
13	10, 11, 12	3bitr4g 223	1 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1364  ∀wral 2475   class class class wbr 4033   Po wpo 4329

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-17 1540  ax-ial 1548  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-cleq 2189  df-clel 2192  df-ral 2480  df-br 4034  df-po 4331

This theorem is referenced by:  soeq1  4350