Theorem poeq2 4011

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 2975	. . 3 ⊢ (A = B → B ⊆ A)
2		poss 4009	. . 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 2974	. . 3 ⊢ (A = B → A ⊆ B)
5		poss 4009	. . 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 120	1 ⊢ (A = B → (𝑅 Po A ↔ 𝑅 Po B))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1228   ⊆ wss 2894   Po wpo 4005

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-11 1378  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-ral 2289  df-in 2901  df-ss 2908  df-po 4007

This theorem is referenced by: (None)