Theorem eu1 2225

Description: An alternate way to express uniqueness used by some authors. Exercise 2(b) of [Margaris] p. 110. (Contributed by NM, 20-Aug-1993.) (Revised by Mario Carneiro, 7-Oct-2016.)

Hypothesis

Ref	Expression
eu1.1	⊢ Ⅎyφ

Assertion

Ref	Expression
eu1	⊢ (∃!xφ ↔ ∃x(φ ∧ ∀y([y / x]φ → x = y)))

Distinct variable group:   x,y

Allowed substitution hints:   φ(x, y)

Proof of Theorem eu1

Step	Hyp	Ref	Expression
1		nfs1v 2106	. . 3 ⊢ Ⅎx[y / x]φ
2	1	euf 2210	. 2 ⊢ (∃!y[y / x]φ ↔ ∃x∀y([y / x]φ ↔ y = x))
3		eu1.1	. . 3 ⊢ Ⅎyφ
4	3	sb8eu 2222	. 2 ⊢ (∃!xφ ↔ ∃!y[y / x]φ)
5		equcom 1680	. . . . . . 7 ⊢ (x = y ↔ y = x)
6	5	imbi2i 303	. . . . . 6 ⊢ (([y / x]φ → x = y) ↔ ([y / x]φ → y = x))
7	6	albii 1566	. . . . 5 ⊢ (∀y([y / x]φ → x = y) ↔ ∀y([y / x]φ → y = x))
8	3	sb6rf 2091	. . . . 5 ⊢ (φ ↔ ∀y(y = x → [y / x]φ))
9	7, 8	anbi12i 678	. . . 4 ⊢ ((∀y([y / x]φ → x = y) ∧ φ) ↔ (∀y([y / x]φ → y = x) ∧ ∀y(y = x → [y / x]φ)))
10		ancom 437	. . . 4 ⊢ ((φ ∧ ∀y([y / x]φ → x = y)) ↔ (∀y([y / x]φ → x = y) ∧ φ))
11		albiim 1611	. . . 4 ⊢ (∀y([y / x]φ ↔ y = x) ↔ (∀y([y / x]φ → y = x) ∧ ∀y(y = x → [y / x]φ)))
12	9, 10, 11	3bitr4i 268	. . 3 ⊢ ((φ ∧ ∀y([y / x]φ → x = y)) ↔ ∀y([y / x]φ ↔ y = x))
13	12	exbii 1582	. 2 ⊢ (∃x(φ ∧ ∀y([y / x]φ → x = y)) ↔ ∃x∀y([y / x]φ ↔ y = x))
14	2, 4, 13	3bitr4i 268	1 ⊢ (∃!xφ ↔ ∃x(φ ∧ ∀y([y / x]φ → x = y)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541  Ⅎwnf 1544   = wceq 1642  [wsb 1648  ∃!weu 2204

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

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-eu 2208

This theorem is used by:  euex  2227  eu2  2229  fvfullfunlem1  5862