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Theorem 3elpr2eq 4908
Description: If there are three elements in a proper unordered pair, and two of them are different from the third one, the two must be equal. (Contributed by AV, 19-Dec-2021.)
Assertion
Ref Expression
3elpr2eq (((𝑋 ∈ {𝐴, 𝐵} ∧ 𝑌 ∈ {𝐴, 𝐵} ∧ 𝑍 ∈ {𝐴, 𝐵}) ∧ (𝑌𝑋𝑍𝑋)) → 𝑌 = 𝑍)

Proof of Theorem 3elpr2eq
StepHypRef Expression
1 elpri 4651 . . 3 (𝑋 ∈ {𝐴, 𝐵} → (𝑋 = 𝐴𝑋 = 𝐵))
2 elpri 4651 . . 3 (𝑌 ∈ {𝐴, 𝐵} → (𝑌 = 𝐴𝑌 = 𝐵))
3 elpri 4651 . . 3 (𝑍 ∈ {𝐴, 𝐵} → (𝑍 = 𝐴𝑍 = 𝐵))
4 eqtr3 2759 . . . . . . . . . . 11 ((𝑍 = 𝐴𝑋 = 𝐴) → 𝑍 = 𝑋)
5 eqneqall 2952 . . . . . . . . . . 11 (𝑍 = 𝑋 → (𝑍𝑋𝑌 = 𝑍))
64, 5syl 17 . . . . . . . . . 10 ((𝑍 = 𝐴𝑋 = 𝐴) → (𝑍𝑋𝑌 = 𝑍))
76adantld 492 . . . . . . . . 9 ((𝑍 = 𝐴𝑋 = 𝐴) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
87ex 414 . . . . . . . 8 (𝑍 = 𝐴 → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
98a1d 25 . . . . . . 7 (𝑍 = 𝐴 → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
10 eqtr3 2759 . . . . . . . . . . . . 13 ((𝑌 = 𝐴𝑋 = 𝐴) → 𝑌 = 𝑋)
11 eqneqall 2952 . . . . . . . . . . . . 13 (𝑌 = 𝑋 → (𝑌𝑋 → (𝑍𝑋𝑌 = 𝑍)))
1210, 11syl 17 . . . . . . . . . . . 12 ((𝑌 = 𝐴𝑋 = 𝐴) → (𝑌𝑋 → (𝑍𝑋𝑌 = 𝑍)))
1312impd 412 . . . . . . . . . . 11 ((𝑌 = 𝐴𝑋 = 𝐴) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
1413ex 414 . . . . . . . . . 10 (𝑌 = 𝐴 → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
1514a1d 25 . . . . . . . . 9 (𝑌 = 𝐴 → (𝑍 = 𝐵 → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
16 eqtr3 2759 . . . . . . . . . . 11 ((𝑌 = 𝐵𝑍 = 𝐵) → 𝑌 = 𝑍)
17162a1d 26 . . . . . . . . . 10 ((𝑌 = 𝐵𝑍 = 𝐵) → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
1817ex 414 . . . . . . . . 9 (𝑌 = 𝐵 → (𝑍 = 𝐵 → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
1915, 18jaoi 856 . . . . . . . 8 ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑍 = 𝐵 → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
2019com12 32 . . . . . . 7 (𝑍 = 𝐵 → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
219, 20jaoi 856 . . . . . 6 ((𝑍 = 𝐴𝑍 = 𝐵) → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐴 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
2221com13 88 . . . . 5 (𝑋 = 𝐴 → ((𝑌 = 𝐴𝑌 = 𝐵) → ((𝑍 = 𝐴𝑍 = 𝐵) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
23 eqtr3 2759 . . . . . . . . . . 11 ((𝑌 = 𝐴𝑍 = 𝐴) → 𝑌 = 𝑍)
24232a1d 26 . . . . . . . . . 10 ((𝑌 = 𝐴𝑍 = 𝐴) → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
2524ex 414 . . . . . . . . 9 (𝑌 = 𝐴 → (𝑍 = 𝐴 → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
26 eqtr3 2759 . . . . . . . . . . . . 13 ((𝑌 = 𝐵𝑋 = 𝐵) → 𝑌 = 𝑋)
2726, 11syl 17 . . . . . . . . . . . 12 ((𝑌 = 𝐵𝑋 = 𝐵) → (𝑌𝑋 → (𝑍𝑋𝑌 = 𝑍)))
2827impd 412 . . . . . . . . . . 11 ((𝑌 = 𝐵𝑋 = 𝐵) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
2928ex 414 . . . . . . . . . 10 (𝑌 = 𝐵 → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
3029a1d 25 . . . . . . . . 9 (𝑌 = 𝐵 → (𝑍 = 𝐴 → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
3125, 30jaoi 856 . . . . . . . 8 ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑍 = 𝐴 → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
3231com12 32 . . . . . . 7 (𝑍 = 𝐴 → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
33 eqtr3 2759 . . . . . . . . . . 11 ((𝑍 = 𝐵𝑋 = 𝐵) → 𝑍 = 𝑋)
3433, 5syl 17 . . . . . . . . . 10 ((𝑍 = 𝐵𝑋 = 𝐵) → (𝑍𝑋𝑌 = 𝑍))
3534adantld 492 . . . . . . . . 9 ((𝑍 = 𝐵𝑋 = 𝐵) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
3635ex 414 . . . . . . . 8 (𝑍 = 𝐵 → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍)))
3736a1d 25 . . . . . . 7 (𝑍 = 𝐵 → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
3832, 37jaoi 856 . . . . . 6 ((𝑍 = 𝐴𝑍 = 𝐵) → ((𝑌 = 𝐴𝑌 = 𝐵) → (𝑋 = 𝐵 → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
3938com13 88 . . . . 5 (𝑋 = 𝐵 → ((𝑌 = 𝐴𝑌 = 𝐵) → ((𝑍 = 𝐴𝑍 = 𝐵) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
4022, 39jaoi 856 . . . 4 ((𝑋 = 𝐴𝑋 = 𝐵) → ((𝑌 = 𝐴𝑌 = 𝐵) → ((𝑍 = 𝐴𝑍 = 𝐵) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))))
41403imp 1112 . . 3 (((𝑋 = 𝐴𝑋 = 𝐵) ∧ (𝑌 = 𝐴𝑌 = 𝐵) ∧ (𝑍 = 𝐴𝑍 = 𝐵)) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
421, 2, 3, 41syl3an 1161 . 2 ((𝑋 ∈ {𝐴, 𝐵} ∧ 𝑌 ∈ {𝐴, 𝐵} ∧ 𝑍 ∈ {𝐴, 𝐵}) → ((𝑌𝑋𝑍𝑋) → 𝑌 = 𝑍))
4342imp 408 1 (((𝑋 ∈ {𝐴, 𝐵} ∧ 𝑌 ∈ {𝐴, 𝐵} ∧ 𝑍 ∈ {𝐴, 𝐵}) ∧ (𝑌𝑋𝑍𝑋)) → 𝑌 = 𝑍)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wa 397  wo 846  w3a 1088   = wceq 1542  wcel 2107  wne 2941  {cpr 4631
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704
This theorem depends on definitions:  df-bi 206  df-an 398  df-or 847  df-3an 1090  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-ne 2942  df-v 3477  df-un 3954  df-sn 4630  df-pr 4632
This theorem is referenced by:  numedglnl  28404
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