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Theorem sban 1829
 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 1769 . . . 4 ([𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
21sbbii 1648 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
3 sbanv 1769 . . 3 ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
42, 3bitri 173 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5 ax-17 1419 . . 3 ((𝜑𝜓) → ∀𝑧(𝜑𝜓))
65sbco2v 1821 . 2 ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑥](𝜑𝜓))
7 ax-17 1419 . . . 4 (𝜑 → ∀𝑧𝜑)
87sbco2v 1821 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9 ax-17 1419 . . . 4 (𝜓 → ∀𝑧𝜓)
109sbco2v 1821 . . 3 ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
118, 10anbi12i 433 . 2 (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
124, 6, 113bitr3i 199 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
 Colors of variables: wff set class Syntax hints:   ∧ wa 97   ↔ wb 98  [wsb 1645 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428 This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646 This theorem is referenced by:  sb3an  1832  sbbi  1833  sbmo  1959  moanim  1974  sbabel  2203  nfrexdya  2359  cbvreu  2531  sbcan  2805  sbcang  2806  rmo3  2849  inab  3205  difab  3206  exss  3963  inopab  4468  bdcriota  10003
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