Theorem sban 2119

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 1903. (Contributed by NM, 14-May-1993.) (Proof shortened by Steven Nguyen, 13-Aug-2023.)

Assertion

Ref	Expression
sban	⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Proof of Theorem sban

Step	Hyp	Ref	Expression
1		simpl 488	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
2	1	sbimi 2113	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜑)
3		simpr 490	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
4	3	sbimi 2113	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜓)
5	2, 4	jca 521	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
6		pm3.2 475	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜑 ∧ 𝜓)))
7	6	sb2imi 2114	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥](𝜑 ∧ 𝜓)))
8	7	imp 412	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 ∧ 𝜓))
9	5, 8	impbii 212	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401  [wsb 2101

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102

This theorem is used by:  sb3an  2120  sbbi  2344  sbabel  2958  cbvreu  3408  rmo3f  3695  sbcan  3791  rmo3  3840  inab  4257  difab  4258  exss  5436  inopab  5814  difopab  5815  mo5f  33160  iuninc  33230  suppss2f  33307  fmptdf2  33325  disjdsct  33370  esumpfinvalf  34791  measiuns  34933  ballotlemodife  35213  xpab  36682  sbn1ALT  37961  sb5ALT  45710  2uasbanh  45746  2uasbanhVD  46095  sb5ALTVD  46097  ellimcabssub0  46844  ichan  48752