Theorem sban 2117

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 2111	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜑)
3		simpr 490	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
4	3	sbimi 2111	. . 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 2112	. . 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 2099

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 1943  ax-6 2000  ax-7 2041

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100

This theorem is used by:  sb3an  2118  sbbi  2345  sbabel  2960  cbvreu  3411  rmo3f  3700  sbcan  3796  rmo3  3845  inab  4265  difab  4266  exss  5449  inopab  5821  difopab  5822  mo5f  32872  iuninc  32942  suppss2f  33020  fmptdf2  33038  disjdsct  33085  esumpfinvalf  34497  measiuns  34639  ballotlemodife  34920  xpab  36239  sbn1ALT  37534  sb5ALT  45275  2uasbanh  45311  2uasbanhVD  45660  sb5ALTVD  45662  ellimcabssub0  46374  ichan  48245