Theorem poeq2 4397

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 3282	. . 3 ⊢ (𝐴 = 𝐵 → 𝐵 ⊆ 𝐴)
2		poss 4395	. . 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 3281	. . 3 ⊢ (𝐴 = 𝐵 → 𝐴 ⊆ 𝐵)
5		poss 4395	. . 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 129	1 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝐴 ↔ 𝑅 Po 𝐵))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1397   ⊆ wss 3200   Po wpo 4391

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-11 1554  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-ral 2515  df-in 3206  df-ss 3213  df-po 4393

This theorem is referenced by: (None)