Theorem poeq2 4028

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 2992	. . 3 ⊢ (A = B → B ⊆ A)
2		poss 4026	. . 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 2991	. . 3 ⊢ (A = B → A ⊆ B)
5		poss 4026	. . 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 1242   ⊆ wss 2911   Po wpo 4022

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 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-ral 2305  df-in 2918  df-ss 2925  df-po 4024

This theorem is referenced by: (None)