Theorem poeq2 5235

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 3852	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		poss 5233	. . 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 3851	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		poss 5233	. . 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 204	1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Syntax hints:   → wi 4   ↔ wb 198   = wceq 1653   ⊆ wss 3767   Po wpo 5229

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-ext 2775

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2784  df-cleq 2790  df-clel 2793  df-ral 3092  df-in 3774  df-ss 3781  df-po 5231

This theorem is referenced by:  posn  5390  frfi  8445  dfpo2  32151  ipo0  39421