Theorem sbel2x 1991

Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbel2x	⊢ (𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))

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

Allowed substitution hints:   𝜑(𝑧,𝑤)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 1990	. . . . 5 ⊢ ([𝑥 / 𝑧]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
2	1	anbi2i 454	. . . 4 ⊢ ((𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
3	2	exbii 1598	. . 3 ⊢ (∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
4		sbelx 1990	. . 3 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑))
5		exdistr 1902	. . 3 ⊢ (∃𝑥∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)) ↔ ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
6	3, 4, 5	3bitr4i 211	. 2 ⊢ (𝜑 ↔ ∃𝑥∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
7		anass 399	. . 3 ⊢ (((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
8	7	2exbii 1599	. 2 ⊢ (∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ∃𝑥∃𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
9	6, 8	bitr4i 186	1 ⊢ (𝜑 ↔ ∃𝑥∃𝑦((𝑥 = 𝑧 ∧ 𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))

Syntax hints:   ∧ wa 103   ↔ wb 104  ∃wex 1485  [wsb 1755

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-sb 1756

This theorem is referenced by: (None)