Theorem sbcan 3028

Description: Distribution of class substitution over conjunction. (Contributed by NM, 31-Dec-2016.)

Assertion

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

Proof of Theorem sbcan

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 2994	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) → 𝐴 ∈ V)
2		sbcex 2994	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
3	2	adantl 277	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
4		dfsbcq2 2988	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∧ 𝜓)))
5		dfsbcq2 2988	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
6		dfsbcq2 2988	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
7	5, 6	anbi12d 473	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
8		sban 1971	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
9	4, 7, 8	vtoclbg 2821	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓)))
10	1, 3, 9	pm5.21nii 705	1 ⊢ ([𝐴 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∧ [𝐴 / 𝑥]𝜓))

Syntax hints:   ∧ wa 104   ↔ wb 105   = wceq 1364  [wsb 1773   ∈ wcel 2164  Vcvv 2760  [wsbc 2985

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-sbc 2986

This theorem is referenced by:  sbc3an  3047  difopab  4795  sbcfung  5278  sbcfng  5401  sbcfg  5402  f1od2  6288