Theorem sban 2011

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 1940	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1814	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
3		sbanv 1940	. . 3 ⊢ ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
4	2, 3	bitri 184	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5		ax-17 1575	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑧(𝜑 ∧ 𝜓))
6	5	sbco2vh 2001	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
7		ax-17 1575	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2vh 2001	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1575	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2vh 2001	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
11	8, 10	anbi12i 460	. 2 ⊢ (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∧ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
12	4, 6, 11	3bitr3i 210	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Syntax hints:   ∧ wa 104   ↔ wb 105  [wsb 1811

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584

This theorem depends on definitions:  df-bi 117  df-nf 1510  df-sb 1812

This theorem is referenced by:  sb3an  2014  sbbi  2015  sbmo  2142  moanim  2157  sbabel  2413  nfrexdya  2580  cbvreu  2778  rmo3f  3016  sbcan  3087  sbcang  3088  rmo3  3137  inab  3491  difab  3492  exss  4345  inopab  4889  bdcriota  16702