Theorem euor2 2272

Description: Introduce or eliminate a disjunct in a uniqueness quantifier. (Contributed by NM, 21-Oct-2005.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)

Assertion

Ref	Expression
euor2	⊢ (¬ ∃xφ → (∃!x(φ ∨ ψ) ↔ ∃!xψ))

Proof of Theorem euor2

Step	Hyp	Ref	Expression
1		nfe1 1732	. . 3 ⊢ Ⅎx∃xφ
2	1	nfn 1793	. 2 ⊢ Ⅎx ¬ ∃xφ
3		19.8a 1756	. . . 4 ⊢ (φ → ∃xφ)
4	3	con3i 127	. . 3 ⊢ (¬ ∃xφ → ¬ φ)
5		orel1 371	. . . 4 ⊢ (¬ φ → ((φ ∨ ψ) → ψ))
6		olc 373	. . . 4 ⊢ (ψ → (φ ∨ ψ))
7	5, 6	impbid1 194	. . 3 ⊢ (¬ φ → ((φ ∨ ψ) ↔ ψ))
8	4, 7	syl 15	. 2 ⊢ (¬ ∃xφ → ((φ ∨ ψ) ↔ ψ))
9	2, 8	eubid 2211	1 ⊢ (¬ ∃xφ → (∃!x(φ ∨ ψ) ↔ ∃!xψ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357  ∃wex 1541  ∃!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-11 1746

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

This theorem is used by:  reuun2  3539