Theorem poeq2 4180

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

Assertion

Ref	Expression
poeq2	⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Proof of Theorem poeq2

Step	Hyp	Ref	Expression
1		eqimss2 3116	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		poss 4178	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵))
3	1, 2	syl 14	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵))
4		eqimss 3115	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		poss 4178	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))
6	4, 5	syl 14	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))
7	3, 6	impbid 128	1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Syntax hints:   → wi 4   ↔ wb 104   = wceq 1312   ⊆ wss 3035   Po wpo 4174

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1404  ax-7 1405  ax-gen 1406  ax-ie1 1450  ax-ie2 1451  ax-8 1463  ax-11 1465  ax-4 1468  ax-17 1487  ax-i9 1491  ax-ial 1495  ax-i5r 1496  ax-ext 2095

This theorem depends on definitions:  df-bi 116  df-nf 1418  df-sb 1717  df-clab 2100  df-cleq 2106  df-clel 2109  df-ral 2393  df-in 3041  df-ss 3048  df-po 4176

This theorem is referenced by: (None)