Theorem reutruALT 50054

Description: Alternate proof of reutru 50053. (Contributed by Zhi Wang, 23-Sep-2024.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
reutruALT	⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

Proof of Theorem reutruALT

Step	Hyp	Ref	Expression
1		rextru 3096	. . 3 ⊢ (∃𝑥 𝑥 ∈ 𝐴 ↔ ∃𝑥 ∈ 𝐴 ⊤)
2		rmotru 50052	. . 3 ⊢ (∃*𝑥 𝑥 ∈ 𝐴 ↔ ∃*𝑥 ∈ 𝐴 ⊤)
3	1, 2	anbi12i 640	. 2 ⊢ ((∃𝑥 𝑥 ∈ 𝐴 ∧ ∃*𝑥 𝑥 ∈ 𝐴) ↔ (∃𝑥 ∈ 𝐴 ⊤ ∧ ∃*𝑥 ∈ 𝐴 ⊤))
4		df-eu 2597	. 2 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ (∃𝑥 𝑥 ∈ 𝐴 ∧ ∃*𝑥 𝑥 ∈ 𝐴))
5		reu5 3370	. 2 ⊢ (∃!𝑥 ∈ 𝐴 ⊤ ↔ (∃𝑥 ∈ 𝐴 ⊤ ∧ ∃*𝑥 ∈ 𝐴 ⊤))
6	3, 4, 5	3bitr4i 306	1 ⊢ (∃!𝑥 𝑥 ∈ 𝐴 ↔ ∃!𝑥 ∈ 𝐴 ⊤)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  ⊤wtru 1570  ∃wex 1811   ∈ wcel 2146  ∃*wmo 2565  ∃!weu 2596  ∃wrex 3089  ∃!wreu 3366  ∃*wrmo 3367

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-mo 2567  df-eu 2597  df-rex 3090  df-rmo 3368  df-reu 3369

This theorem is used by: (None)