Theorem sbex 2319

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 2318	. . 3 ⊢ ([𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑 ↔ ¬ [𝑧 / 𝑦]∀𝑥 ¬ 𝜑)
2		sbn 2318	. . . 4 ⊢ ([𝑧 / 𝑦] ¬ 𝜑 ↔ ¬ [𝑧 / 𝑦]𝜑)
3	2	sbalv 2208	. . 3 ⊢ ([𝑧 / 𝑦]∀𝑥 ¬ 𝜑 ↔ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
4	1, 3	xchbinx 337	. 2 ⊢ ([𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑 ↔ ¬ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
5		df-ex 1813	. . 3 ⊢ (∃𝑥𝜑 ↔ ¬ ∀𝑥 ¬ 𝜑)
6	5	sbbii 2113	. 2 ⊢ ([𝑧 / 𝑦]∃𝑥𝜑 ↔ [𝑧 / 𝑦] ¬ ∀𝑥 ¬ 𝜑)
7		df-ex 1813	. 2 ⊢ (∃𝑥[𝑧 / 𝑦]𝜑 ↔ ¬ ∀𝑥 ¬ [𝑧 / 𝑦]𝜑)
8	4, 6, 7	3bitr4i 306	1 ⊢ ([𝑧 / 𝑦]∃𝑥𝜑 ↔ ∃𝑥[𝑧 / 𝑦]𝜑)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  ∀wal 1568  ∃wex 1812  [wsb 2099

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2179  ax-11 2195  ax-12 2216

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100

This theorem is used by:  sbnf  2349  sbmo  2645  sbabel  2960  sbcex2  3807