Theorem elpreq 32949

Description: Equality wihin a pair. (Contributed by Thierry Arnoux, 23-Aug-2017.)

Hypotheses

Ref	Expression
elpreq.1	⊢ (𝜑 → 𝑋 ∈ {𝐴, 𝐵})
elpreq.2	⊢ (𝜑 → 𝑌 ∈ {𝐴, 𝐵})
elpreq.3	⊢ (𝜑 → (𝑋 = 𝐴 ↔ 𝑌 = 𝐴))

Assertion

Ref	Expression
elpreq	⊢ (𝜑 → 𝑋 = 𝑌)

Proof of Theorem elpreq

Step	Hyp	Ref	Expression
1		simpr 490	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝐴) → 𝑋 = 𝐴)
2		elpreq.3	. . . 4 ⊢ (𝜑 → (𝑋 = 𝐴 ↔ 𝑌 = 𝐴))
3	2	biimpa 482	. . 3 ⊢ ((𝜑 ∧ 𝑋 = 𝐴) → 𝑌 = 𝐴)
4	1, 3	eqtr4d 2803	. 2 ⊢ ((𝜑 ∧ 𝑋 = 𝐴) → 𝑋 = 𝑌)
5		elpreq.1	. . . . 5 ⊢ (𝜑 → 𝑋 ∈ {𝐴, 𝐵})
6		elpri 4615	. . . . 5 ⊢ (𝑋 ∈ {𝐴, 𝐵} → (𝑋 = 𝐴 ∨ 𝑋 = 𝐵))
7	5, 6	syl 18	. . . 4 ⊢ (𝜑 → (𝑋 = 𝐴 ∨ 𝑋 = 𝐵))
8	7	orcanai 1018	. . 3 ⊢ ((𝜑 ∧ ¬ 𝑋 = 𝐴) → 𝑋 = 𝐵)
9		simpl 488	. . . 4 ⊢ ((𝜑 ∧ ¬ 𝑋 = 𝐴) → 𝜑)
10	2	notbid 321	. . . . 5 ⊢ (𝜑 → (¬ 𝑋 = 𝐴 ↔ ¬ 𝑌 = 𝐴))
11	10	biimpa 482	. . . 4 ⊢ ((𝜑 ∧ ¬ 𝑋 = 𝐴) → ¬ 𝑌 = 𝐴)
12		elpreq.2	. . . . 5 ⊢ (𝜑 → 𝑌 ∈ {𝐴, 𝐵})
13		elpri 4615	. . . . 5 ⊢ (𝑌 ∈ {𝐴, 𝐵} → (𝑌 = 𝐴 ∨ 𝑌 = 𝐵))
14		pm2.53 865	. . . . 5 ⊢ ((𝑌 = 𝐴 ∨ 𝑌 = 𝐵) → (¬ 𝑌 = 𝐴 → 𝑌 = 𝐵))
15	12, 13, 14	3syl 19	. . . 4 ⊢ (𝜑 → (¬ 𝑌 = 𝐴 → 𝑌 = 𝐵))
16	9, 11, 15	sylc 66	. . 3 ⊢ ((𝜑 ∧ ¬ 𝑋 = 𝐴) → 𝑌 = 𝐵)
17	8, 16	eqtr4d 2803	. 2 ⊢ ((𝜑 ∧ ¬ 𝑋 = 𝐴) → 𝑋 = 𝑌)
18	4, 17	pm2.61dan 825	1 ⊢ (𝜑 → 𝑋 = 𝑌)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2146  {cpr 4593

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459  df-un 3911  df-sn 4592  df-pr 4594

This theorem is used by:  indpreima  33259