Theorem sbc2rex 43890

Description: Exchange a substitution with two existentials. (Contributed by Stefan O'Rear, 11-Oct-2014.) (Revised by NM, 24-Aug-2018.)

Assertion

Ref	Expression
sbc2rex	⊢ ([𝐴 / 𝑎]∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 𝜑 ↔ ∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 [𝐴 / 𝑎]𝜑)

Distinct variable groups:   𝐴,𝑏   𝐴,𝑐   𝐵,𝑎   𝐶,𝑎   𝑎,𝑏   𝑎,𝑐

Allowed substitution hints:   𝜑(𝑎, 𝑏, 𝑐)   𝐴(𝑎)   𝐵(𝑏, 𝑐)   𝐶(𝑏, 𝑐)

Proof of Theorem sbc2rex

Step	Hyp	Ref	Expression
1		sbcrex 3821	. 2 ⊢ ([𝐴 / 𝑎]∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 𝜑 ↔ ∃𝑏 ∈ 𝐵 [𝐴 / 𝑎]∃𝑐 ∈ 𝐶 𝜑)
2		sbcrex 3821	. . 3 ⊢ ([𝐴 / 𝑎]∃𝑐 ∈ 𝐶 𝜑 ↔ ∃𝑐 ∈ 𝐶 [𝐴 / 𝑎]𝜑)
3	2	rexbii 3109	. 2 ⊢ (∃𝑏 ∈ 𝐵 [𝐴 / 𝑎]∃𝑐 ∈ 𝐶 𝜑 ↔ ∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 [𝐴 / 𝑎]𝜑)
4	1, 3	bitri 278	1 ⊢ ([𝐴 / 𝑎]∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 𝜑 ↔ ∃𝑏 ∈ 𝐵 ∃𝑐 ∈ 𝐶 [𝐴 / 𝑎]𝜑)

This proof depends on syntax axioms:   ↔ wb 209  ∃wrex 3086  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rex 3087  df-v 3452  df-sbc 3739

This theorem is used by:  sbc4rex  43891  3rexfrabdioph  43898  4rexfrabdioph  43899  7rexfrabdioph  43901