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Theorem sban 1967
Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 3-Feb-2018.)
Assertion
Ref Expression
sban ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Proof of Theorem sban
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 sbanv 1901 . . . 4 ([𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
21sbbii 1776 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
3 sbanv 1901 . . 3 ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
42, 3bitri 184 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5 ax-17 1537 . . 3 ((𝜑𝜓) → ∀𝑧(𝜑𝜓))
65sbco2vh 1957 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑥](𝜑𝜓))
7 ax-17 1537 . . . 4 (𝜑 → ∀𝑧𝜑)
87sbco2vh 1957 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9 ax-17 1537 . . . 4 (𝜓 → ∀𝑧𝜓)
109sbco2vh 1957 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
118, 10anbi12i 460 . 2 (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
124, 6, 113bitr3i 210 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
Colors of variables: wff set class
Syntax hints:  wa 104  wb 105  [wsb 1773
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546
This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774
This theorem is referenced by:  sb3an  1970  sbbi  1971  sbmo  2097  moanim  2112  sbabel  2359  nfrexdya  2526  cbvreu  2716  rmo3f  2949  sbcan  3020  sbcang  3021  rmo3  3069  inab  3418  difab  3419  exss  4245  inopab  4777  bdcriota  15093
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