Theorem poeq2 5595

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 4042	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		poss 5593	. . 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 4041	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		poss 5593	. . 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 212	1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Syntax hints:   → wi 4   ↔ wb 206   = wceq 1539   ⊆ wss 3950   Po wpo 5589

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-9 2117  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-cleq 2728  df-ral 3061  df-ss 3967  df-po 5591

This theorem is referenced by:  poeq12d  5596  posn  5770  dfpo2  6315  frfi  9322  ipo0  44473