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Theorem sbel2x 1971
Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)
Assertion
Ref Expression
sbel2x (𝜑 ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 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
StepHypRef Expression
1 sbelx 1970 . . . . 5 ([𝑥 / 𝑧]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
21anbi2i 452 . . . 4 ((𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
32exbii 1584 . . 3 (∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
4 sbelx 1970 . . 3 (𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑))
5 exdistr 1881 . . 3 (∃𝑥𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)) ↔ ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
63, 4, 53bitr4i 211 . 2 (𝜑 ↔ ∃𝑥𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
7 anass 398 . . 3 (((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
872exbii 1585 . 2 (∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ∃𝑥𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
96, 8bitr4i 186 1 (𝜑 ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104  wex 1468  [wsb 1735
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 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-11 1484  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514
This theorem depends on definitions:  df-bi 116  df-sb 1736
This theorem is referenced by: (None)
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