Theorem sbex 2128

Description: Move existential quantifier in and out of substitution. (Contributed by NM, 27-Sep-2003.)

Assertion

Ref	Expression
sbex	⊢ ([z / y]∃xφ ↔ ∃x[z / y]φ)

Distinct variable groups:   x,y   x,z

Allowed substitution hints:   φ(x, y, z)

Proof of Theorem sbex

Step	Hyp	Ref	Expression
1		sbn 2062	. . 3 ⊢ ([z / y] ¬ ∀x ¬ φ ↔ ¬ [z / y]∀x ¬ φ)
2		sbal 2127	. . . 4 ⊢ ([z / y]∀x ¬ φ ↔ ∀x[z / y] ¬ φ)
3		sbn 2062	. . . . 5 ⊢ ([z / y] ¬ φ ↔ ¬ [z / y]φ)
4	3	albii 1566	. . . 4 ⊢ (∀x[z / y] ¬ φ ↔ ∀x ¬ [z / y]φ)
5	2, 4	bitri 240	. . 3 ⊢ ([z / y]∀x ¬ φ ↔ ∀x ¬ [z / y]φ)
6	1, 5	xchbinx 301	. 2 ⊢ ([z / y] ¬ ∀x ¬ φ ↔ ¬ ∀x ¬ [z / y]φ)
7		df-ex 1542	. . 3 ⊢ (∃xφ ↔ ¬ ∀x ¬ φ)
8	7	sbbii 1653	. 2 ⊢ ([z / y]∃xφ ↔ [z / y] ¬ ∀x ¬ φ)
9		df-ex 1542	. 2 ⊢ (∃x[z / y]φ ↔ ¬ ∀x ¬ [z / y]φ)
10	6, 8, 9	3bitr4i 268	1 ⊢ ([z / y]∃xφ ↔ ∃x[z / y]φ)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 176  ∀wal 1540  ∃wex 1541  [wsb 1648

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-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is used by:  sbmo  2234  sbabel  2516  sbcex2  3096  sbcexg  3097