Theorem poeq2 5507

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 3978	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		poss 5505	. . 3 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵))
3	1, 2	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 → 𝑅 Po 𝐵))
4		eqimss 3977	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		poss 5505	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))
6	4, 5	syl 17	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))
7	3, 6	impbid 211	1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1539   ⊆ wss 3887   Po wpo 5501

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-v 3434  df-in 3894  df-ss 3904  df-po 5503

This theorem is referenced by:  posn  5672  dfpo2  6199  frfi  9059  ipo0  42067