Theorem reiota2 4369

Description: A condition allowing us to represent "the unique element in A such that φ " with a class expression B. (Contributed by Scott Fenton, 7-Jan-2018.)

Hypothesis

Ref	Expression
reiota2.1	⊢ (x = B → (φ ↔ ψ))

Assertion

Ref	Expression
reiota2	⊢ ((B ∈ A ∧ ∃!x ∈ A φ) → (ψ ↔ (℩x(x ∈ A ∧ φ)) = B))

Distinct variable groups:   x,A   x,B   ψ,x

Allowed substitution hint:   φ(x)

Proof of Theorem reiota2

Step	Hyp	Ref	Expression
1		simpl 443	. . 3 ⊢ ((B ∈ A ∧ ∃!x ∈ A φ) → B ∈ A)
2	1	biantrurd 494	. 2 ⊢ ((B ∈ A ∧ ∃!x ∈ A φ) → (ψ ↔ (B ∈ A ∧ ψ)))
3		df-reu 2622	. . 3 ⊢ (∃!x ∈ A φ ↔ ∃!x(x ∈ A ∧ φ))
4		eleq1 2413	. . . . 5 ⊢ (x = B → (x ∈ A ↔ B ∈ A))
5		reiota2.1	. . . . 5 ⊢ (x = B → (φ ↔ ψ))
6	4, 5	anbi12d 691	. . . 4 ⊢ (x = B → ((x ∈ A ∧ φ) ↔ (B ∈ A ∧ ψ)))
7	6	iota2 4368	. . 3 ⊢ ((B ∈ A ∧ ∃!x(x ∈ A ∧ φ)) → ((B ∈ A ∧ ψ) ↔ (℩x(x ∈ A ∧ φ)) = B))
8	3, 7	sylan2b 461	. 2 ⊢ ((B ∈ A ∧ ∃!x ∈ A φ) → ((B ∈ A ∧ ψ) ↔ (℩x(x ∈ A ∧ φ)) = B))
9	2, 8	bitrd 244	1 ⊢ ((B ∈ A ∧ ∃!x ∈ A φ) → (ψ ↔ (℩x(x ∈ A ∧ φ)) = B))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642   ∈ wcel 1710  ∃!weu 2204  ∃!wreu 2617  ℩cio 4338

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3an 936  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-eu 2208  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ral 2620  df-rex 2621  df-reu 2622  df-v 2862  df-sbc 3048  df-nin 3212  df-compl 3213  df-un 3215  df-sn 3742  df-pr 3743  df-uni 3893  df-iota 4340

This theorem is used by:  ncfinprop  4475  tfinprop  4490  eqtc  6162