Theorem soeq2 4152

Description: Equality theorem for the strict ordering predicate. (Contributed by NM, 16-Mar-1997.)

Assertion

Ref	Expression
soeq2	⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐴 ↔ 𝑅 Or 𝐵))

Proof of Theorem soeq2

Step	Hyp	Ref	Expression
1		soss 4150	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Or 𝐵 → 𝑅 Or 𝐴))
2		soss 4150	. . . 4 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Or 𝐴 → 𝑅 Or 𝐵))
3	1, 2	anim12i 332	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
4		eqss 3041	. . 3 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		dfbi2 381	. . 3 ⊢ ((𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴) ↔ ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
6	3, 4, 5	3imtr4i 200	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴))
7	6	bicomd 140	1 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐴 ↔ 𝑅 Or 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1290   ⊆ wss 3000   Or wor 4131

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 666  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-10 1442  ax-11 1443  ax-i12 1444  ax-bndl 1445  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-3an 927  df-tru 1293  df-nf 1396  df-sb 1694  df-clab 2076  df-cleq 2082  df-clel 2085  df-nfc 2218  df-ral 2365  df-in 3006  df-ss 3013  df-po 4132  df-iso 4133

This theorem is referenced by: (None)