Theorem sban 1986

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 1916	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1791	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓))
3		sbanv 1916	. . 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 1552	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑧(𝜑 ∧ 𝜓))
6	5	sbco2vh 1976	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓))
7		ax-17 1552	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2vh 1976	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1552	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2vh 1976	. . 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 1788

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 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561

This theorem depends on definitions:  df-bi 117  df-nf 1487  df-sb 1789

This theorem is referenced by:  sb3an  1989  sbbi  1990  sbmo  2117  moanim  2132  sbabel  2379  nfrexdya  2546  cbvreu  2743  rmo3f  2980  sbcan  3051  sbcang  3052  rmo3  3101  inab  3452  difab  3453  exss  4292  inopab  4831  bdcriota  16156