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Theorem sban 1845
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 1785 . . . 4 ([𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
21sbbii 1664 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
3 sbanv 1785 . . 3 ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
42, 3bitri 177 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5 ax-17 1435 . . 3 ((𝜑𝜓) → ∀𝑧(𝜑𝜓))
65sbco2v 1837 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑥](𝜑𝜓))
7 ax-17 1435 . . . 4 (𝜑 → ∀𝑧𝜑)
87sbco2v 1837 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9 ax-17 1435 . . . 4 (𝜓 → ∀𝑧𝜓)
109sbco2v 1837 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
118, 10anbi12i 441 . 2 (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
124, 6, 113bitr3i 203 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
Colors of variables: wff set class
Syntax hints:  wa 101  wb 102  [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-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444
This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662
This theorem is referenced by:  sb3an  1848  sbbi  1849  sbmo  1975  moanim  1990  sbabel  2219  nfrexdya  2376  cbvreu  2548  sbcan  2827  sbcang  2828  rmo3  2876  inab  3232  difab  3233  exss  3990  inopab  4495  bdcriota  10362
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