Theorem 2sb5 1900

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 1808	. 2 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑))
2		19.42v 1827	. . . 4 ⊢ (∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
3		anass 393	. . . . 5 ⊢ (((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)))
4	3	exbii 1536	. . . 4 ⊢ (∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑) ↔ ∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)))
5		sb5 1808	. . . . 5 ⊢ ([𝑤 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑))
6	5	anbi2i 444	. . . 4 ⊢ ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
7	2, 4, 6	3bitr4ri 211	. . 3 ⊢ ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))
8	7	exbii 1536	. 2 ⊢ (∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))
9	1, 8	bitri 182	1 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))

Syntax hints:   ∧ wa 102   ↔ wb 103  ∃wex 1421  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  opelopabsbALT  4014