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Theorem sbel2x 1890
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 1889 . . . . 5 ([𝑥 / 𝑧]𝜑 ↔ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
21anbi2i 438 . . . 4 ((𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
32exbii 1512 . . 3 (∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑) ↔ ∃𝑥(𝑥 = 𝑧 ∧ ∃𝑦(𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
4 sbelx 1889 . . 3 (𝜑 ↔ ∃𝑥(𝑥 = 𝑧 ∧ [𝑥 / 𝑧]𝜑))
5 exdistr 1803 . . 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 205 . 2 (𝜑 ↔ ∃𝑥𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
7 anass 387 . . 3 (((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ (𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
872exbii 1513 . 2 (∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑) ↔ ∃𝑥𝑦(𝑥 = 𝑧 ∧ (𝑦 = 𝑤 ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑)))
96, 8bitr4i 180 1 (𝜑 ↔ ∃𝑥𝑦((𝑥 = 𝑧𝑦 = 𝑤) ∧ [𝑦 / 𝑤][𝑥 / 𝑧]𝜑))
Colors of variables: wff set class
Syntax hints:  wa 101  wb 102  wex 1397  [wsb 1661
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-11 1413  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443
This theorem depends on definitions:  df-bi 114  df-sb 1662
This theorem is referenced by: (None)
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