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](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Proof of Theorem sban

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbanv 1769	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1648	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [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))
4	2, 3	bitri 173	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5		ax-17 1419	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑧(𝜑 ∧ 𝜓))
6	5	sbco2v 1821	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
7		ax-17 1419	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2v 1821	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1419	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2v 1821	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
11	8, 10	anbi12i 433	. 2 ⊢ (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
12	4, 6, 11	3bitr3i 199	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

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