Theorem sban 2087

Description: Conjunction inside and outside of a substitution are equivalent. Compare 19.26 1877. (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 483	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
2	1	sbimi 2081	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜑)
3		simpr 485	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜓)
4	3	sbimi 2081	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → [𝑦 / 𝑥]𝜓)
5	2, 4	jca 512	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) → ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
6		pm3.2 470	. . . 4 ⊢ (𝜑 → (𝜓 → (𝜑 ∧ 𝜓)))
7	6	sb2imi 2082	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥](𝜑 ∧ 𝜓)))
8	7	imp 407	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) → [𝑦 / 𝑥](𝜑 ∧ 𝜓))
9	5, 8	impbii 208	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Syntax hints:   ↔ wb 205   ∧ wa 396  [wsb 2071

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816

This theorem depends on definitions:  df-bi 206  df-an 397  df-sb 2072

This theorem is referenced by:  sb3an  2088  sbbi  2309  sbabel  2943  sbabelOLD  2944  cbvreuw  3374  cbvreu  3379  rmo3f  3673  sbcan  3772  rmo3  3827  inab  4239  difab  4240  exss  5382  inopab  5738  mo5f  30846  iuninc  30909  suppss2f  30983  fmptdF  31002  disjdsct  31044  esumpfinvalf  32053  measiuns  32194  ballotlemodife  32473  xpab  33686  sbn1ALT  35051  sb5ALT  42127  2uasbanh  42163  2uasbanhVD  42513  sb5ALTVD  42515  ellimcabssub0  43140  ichan  44886