Theorem poeq1 4221

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 3931	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑥 ↔ 𝑥𝑆𝑥))
2	1	notbid 656	. . . . 5 ⊢ (𝑅 = 𝑆 → (¬ 𝑥𝑅𝑥 ↔ ¬ 𝑥𝑆𝑥))
3		breq 3931	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑦 ↔ 𝑥𝑆𝑦))
4		breq 3931	. . . . . . 7 ⊢ (𝑅 = 𝑆 → (𝑦𝑅𝑧 ↔ 𝑦𝑆𝑧))
5	3, 4	anbi12d 464	. . . . . 6 ⊢ (𝑅 = 𝑆 → ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) ↔ (𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧)))
6		breq 3931	. . . . . 6 ⊢ (𝑅 = 𝑆 → (𝑥𝑅𝑧 ↔ 𝑥𝑆𝑧))
7	5, 6	imbi12d 233	. . . . 5 ⊢ (𝑅 = 𝑆 → (((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧) ↔ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
8	2, 7	anbi12d 464	. . . 4 ⊢ (𝑅 = 𝑆 → ((¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
9	8	ralbidv 2437	. . 3 ⊢ (𝑅 = 𝑆 → (∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) ↔ ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧))))
10	9	2ralbidv 2459	. 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 4218	. 2 ⊢ (𝑅 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
12		df-po 4218	. 2 ⊢ (𝑆 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑆𝑥 ∧ ((𝑥𝑆𝑦 ∧ 𝑦𝑆𝑧) → 𝑥𝑆𝑧)))
13	10, 11, 12	3bitr4g 222	1 ⊢ (𝑅 = 𝑆 → (𝑅 Po 𝐴 ↔ 𝑆 Po 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1331  ∀wral 2416   class class class wbr 3929   Po wpo 4216

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-4 1487  ax-17 1506  ax-ial 1514  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-nf 1437  df-cleq 2132  df-clel 2135  df-ral 2421  df-br 3930  df-po 4218

This theorem is referenced by:  soeq1  4237