Theorem 2sb5 2282

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 2276	. 2 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑))
2		19.42v 1954	. . . 4 ⊢ (∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ 𝜑)) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
3		anass 471	. . . . 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 2276	. . . . 5 ⊢ ([𝑤 / 𝑦]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑))
6	5	anbi2i 624	. . . 4 ⊢ ((𝑥 = 𝑧 ∧ [𝑤 / 𝑦]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ 𝜑)))
7	2, 4, 6	3bitr4ri 306	. . 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 277	1 ⊢ ([𝑧 / 𝑥][𝑤 / 𝑦]𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ 𝜑))

Syntax hints:   ↔ wb 208   ∧ wa 398  ∃wex 1780  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2145  ax-12 2177

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1781  df-nf 1785  df-sb 2070

This theorem is referenced by:  opelopabsbALT  5418