Theorem sbccomieg 37864

Description: Commute two explicit substitutions, using an implicit substitution to rewrite the exiting substitution. (Contributed by Stefan O'Rear, 11-Oct-2014.) (Revised by Mario Carneiro, 11-Dec-2016.)

Hypothesis

Ref	Expression
sbccomieg.1	⊢ (𝑎 = 𝐴 → 𝐵 = 𝐶)

Assertion

Ref	Expression
sbccomieg	⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑)

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

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

Proof of Theorem sbccomieg

Step	Hyp	Ref	Expression
1		sbcex 3654	. 2 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 → 𝐴 ∈ V)
2		spesbc 3727	. . 3 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → ∃𝑏[𝐴 / 𝑎]𝜑)
3		sbcex 3654	. . . 4 ⊢ ([𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
4	3	exlimiv 2021	. . 3 ⊢ (∃𝑏[𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
5	2, 4	syl 17	. 2 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
6		nfcv 2959	. . . 4 ⊢ Ⅎ𝑎𝐶
7		nfsbc1v 3664	. . . 4 ⊢ Ⅎ𝑎[𝐴 / 𝑎]𝜑
8	6, 7	nfsbc 3666	. . 3 ⊢ Ⅎ𝑎[𝐶 / 𝑏][𝐴 / 𝑎]𝜑
9		sbccomieg.1	. . . 4 ⊢ (𝑎 = 𝐴 → 𝐵 = 𝐶)
10		sbceq1a 3655	. . . 4 ⊢ (𝑎 = 𝐴 → (𝜑 ↔ [𝐴 / 𝑎]𝜑))
11	9, 10	sbceqbid 3651	. . 3 ⊢ (𝑎 = 𝐴 → ([𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
12	8, 11	sbciegf 3676	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
13	1, 5, 12	pm5.21nii 369	1 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑)

Syntax hints:   → wi 4   ↔ wb 197   = wceq 1637  ∃wex 1859   ∈ wcel 2157  Vcvv 3402  [wsbc 3644

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2069  ax-7 2105  ax-9 2166  ax-10 2186  ax-11 2202  ax-12 2215  ax-13 2422  ax-ext 2795

This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-3an 1102  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2062  df-clab 2804  df-cleq 2810  df-clel 2813  df-nfc 2948  df-ral 3112  df-rex 3113  df-v 3404  df-sbc 3645

This theorem is referenced by:  2rexfrabdioph  37867  3rexfrabdioph  37868  4rexfrabdioph  37869  6rexfrabdioph  37870  7rexfrabdioph  37871