Theorem soeq2 4368

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 4366	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Or 𝐵 → 𝑅 Or 𝐴))
2		soss 4366	. . . 4 ⊢ (𝐵 ⊆ 𝐴 → (𝑅 Or 𝐴 → 𝑅 Or 𝐵))
3	1, 2	anim12i 338	. . 3 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴) → ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
4		eqss 3210	. . 3 ⊢ (𝐴 = 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ 𝐴))
5		dfbi2 388	. . 3 ⊢ ((𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴) ↔ ((𝑅 Or 𝐵 → 𝑅 Or 𝐴) ∧ (𝑅 Or 𝐴 → 𝑅 Or 𝐵)))
6	3, 4, 5	3imtr4i 201	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐵 ↔ 𝑅 Or 𝐴))
7	6	bicomd 141	1 ⊢ (𝐴 = 𝐵 → (𝑅 Or 𝐴 ↔ 𝑅 Or 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1373   ⊆ wss 3168   Or wor 4347

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-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ral 2490  df-in 3174  df-ss 3181  df-po 4348  df-iso 4349

This theorem is referenced by: (None)