Theorem sban 2080

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 1870. (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 482	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
2	1	sbimi 2074	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜑)
3		simpr 484	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
4	3	sbimi 2074	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜓)
5	2, 4	jca 511	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
6		pm3.2 469	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜑 ∧ 𝜓)))
7	6	sb2imi 2075	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥](𝜑 ∧ 𝜓)))
8	7	imp 406	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 ∧ 𝜓))
9	5, 8	impbii 209	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Syntax hints:   ↔ wb 206   ∧ wa 395  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396  df-sb 2065

This theorem is referenced by:  sb3an  2081  sbbi  2308  sbabel  2931  cbvreuwOLD  3394  cbvreu  3407  rmo3f  3717  sbcan  3815  rmo3  3864  inab  4284  difab  4285  exss  5438  inopab  5808  difopab  5809  mo5f  32470  iuninc  32541  suppss2f  32616  fmptdF  32634  disjdsct  32680  esumpfinvalf  34107  measiuns  34248  ballotlemodife  34530  xpab  35743  sbn1ALT  36876  sb5ALT  44550  2uasbanh  44586  2uasbanhVD  44935  sb5ALTVD  44937  ellimcabssub0  45646  ichan  47469