Theorem reuss 3537

Description: Transfer uniqueness to a smaller subclass. (Contributed by NM, 21-Aug-1999.)

Assertion

Ref	Expression
reuss	⊢ ((A ⊆ B ∧ ∃x ∈ A φ ∧ ∃!x ∈ B φ) → ∃!x ∈ A φ)

Distinct variable groups:   x,A   x,B

Allowed substitution hint:   φ(x)

Proof of Theorem reuss

Step	Hyp	Ref	Expression
1		idd 21	. . . 4 ⊢ (x ∈ A → (φ → φ))
2	1	rgen 2680	. . 3 ⊢ ∀x ∈ A (φ → φ)
3		reuss2 3536	. . 3 ⊢ (((A ⊆ B ∧ ∀x ∈ A (φ → φ)) ∧ (∃x ∈ A φ ∧ ∃!x ∈ B φ)) → ∃!x ∈ A φ)
4	2, 3	mpanl2 662	. 2 ⊢ ((A ⊆ B ∧ (∃x ∈ A φ ∧ ∃!x ∈ B φ)) → ∃!x ∈ A φ)
5	4	3impb 1147	1 ⊢ ((A ⊆ B ∧ ∃x ∈ A φ ∧ ∃!x ∈ B φ) → ∃!x ∈ A φ)

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   ∧ w3a 934   ∈ wcel 1710  ∀wral 2615  ∃wrex 2616  ∃!wreu 2617   ⊆ wss 3258

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-mo 2209  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-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is used by: (None)