Theorem sbc3an 3806

Description: Distribution of class substitution over triple conjunction. (Contributed by NM, 14-Dec-2006.) (Revised by NM, 17-Aug-2018.)

Assertion

Ref	Expression
sbc3an	⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbc3an

Step	Hyp	Ref	Expression
1		sbcan 3791	. . 3 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝐴 / 𝑥]𝜒))
2		sbcan 3791	. . 3 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))
3	1, 2	bianbi 639	. 2 ⊢ ([𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
4		df-3an 1105	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
5	4	sbcbii 3798	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝐴 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
6		df-3an 1105	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒) ↔ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) ∧ [𝐴 / 𝑥]𝜒))
7	3, 5, 6	3bitr4i 306	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓 ∧ [𝐴 / 𝑥]𝜒))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∧ w3a 1103  [wsbc 3742

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  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-sbc 3743

This theorem is used by:  csbfrecsg  8287  bnj156  35246  bnj206  35249  bnj976  35295  bnj121  35387  bnj130  35391  bnj581  35425  bnj1040  35489  topdifinffinlem  38109  rdgeqoa  38132  cdlemkid3N  41814  cdlemkid4  41815  minregex  44382