Theorem soeq2 4071

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 4069	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Or 𝐵 → 𝑅 Or 𝐴))
2		soss 4069	. . . 4 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Or 𝐴 → 𝑅 Or 𝐵))
3	1, 2	anim12i 331	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
4		eqss 3014	. . 3 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		dfbi2 380	. . 3 ⊢ ((𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴) ↔ ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
6	3, 4, 5	3imtr4i 199	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴))
7	6	bicomd 139	1 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐴 ↔ 𝑅 Or 𝐵))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284   ⊆ wss 2973   Or wor 4050

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-in 2979  df-ss 2986  df-po 4051  df-iso 4052

This theorem is referenced by: (None)