Theorem 2sb5 2279

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)

Assertion

Ref	Expression
2sb5	⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))

Distinct variable groups:   𝑥,𝑦,𝑧   𝑦,𝑤

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

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 2277	. 2 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑))
2		19.42v 1953	. . . 4 ⊢ (∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
3		anass 468	. . . . 5 ⊢ (((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)))
4	3	exbii 1848	. . . 4 ⊢ (∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑) ↔ ∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)))
5		sb5 2277	. . . . 5 ⊢ ([𝑤 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑))
6	5	anbi2i 623	. . . 4 ⊢ ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
7	2, 4, 6	3bitr4ri 304	. . 3 ⊢ ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))
8	7	exbii 1848	. 2 ⊢ (∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))
9	1, 8	bitri 275	1 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))

Syntax hints:   ↔ wb 206   ∧ wa 395  ∃wex 1779  [wsb 2065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-10 2142  ax-12 2178

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-nf 1784  df-sb 2066

This theorem is referenced by:  vopelopabsb  5509