Theorem po2ne 5488

Description: Two classes which are in a partial order relation are not equal. (Contributed by AV, 13-Mar-2023.)

Assertion

Ref	Expression
po2ne	⊢ ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) ∧ 𝐴𝑅𝐵) → 𝐴 ≠ 𝐵)

Proof of Theorem po2ne

Step	Hyp	Ref	Expression
1		breq1 5068	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴𝑅𝐵 ↔ 𝐵𝑅𝐵))
2		poirr 5484	. . . . . . . . 9 ⊢ ((𝑅 Po 𝑉 ∧ 𝐵 ∈ 𝑉) → ¬ 𝐵𝑅𝐵)
3	2	adantrl 714	. . . . . . . 8 ⊢ ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉)) → ¬ 𝐵𝑅𝐵)
4	3	pm2.21d 121	. . . . . . 7 ⊢ ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉)) → (𝐵𝑅𝐵 → 𝐴 ≠ 𝐵))
5	4	ex 415	. . . . . 6 ⊢ (𝑅 Po 𝑉 → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) → (𝐵𝑅𝐵 → 𝐴 ≠ 𝐵)))
6	5	com13 88	. . . . 5 ⊢ (𝐵𝑅𝐵 → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) → (𝑅 Po 𝑉 → 𝐴 ≠ 𝐵)))
7	1, 6	syl6bi 255	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴𝑅𝐵 → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) → (𝑅 Po 𝑉 → 𝐴 ≠ 𝐵))))
8	7	com24 95	. . 3 ⊢ (𝐴 = 𝐵 → (𝑅 Po 𝑉 → ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) → (𝐴𝑅𝐵 → 𝐴 ≠ 𝐵))))
9	8	3impd 1344	. 2 ⊢ (𝐴 = 𝐵 → ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) ∧ 𝐴𝑅𝐵) → 𝐴 ≠ 𝐵))
10		ax-1 6	. 2 ⊢ (𝐴 ≠ 𝐵 → ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) ∧ 𝐴𝑅𝐵) → 𝐴 ≠ 𝐵))
11	9, 10	pm2.61ine 3100	1 ⊢ ((𝑅 Po 𝑉 ∧ (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑉) ∧ 𝐴𝑅𝐵) → 𝐴 ≠ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 398   ∧ w3a 1083   = wceq 1533   ∈ wcel 2110   ≠ wne 3016   class class class wbr 5065   Po wpo 5471

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ne 3017  df-ral 3143  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4567  df-pr 4569  df-op 4573  df-br 5066  df-po 5473

This theorem is referenced by:  prproropf1olem1  43664