Theorem sbccomieg 41834

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 3787	. 2 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 → 𝐴 ∈ V)
2		spesbc 3876	. . 3 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → ∃𝑏[𝐴 / 𝑎]𝜑)
3		sbcex 3787	. . . 4 ⊢ ([𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
4	3	exlimiv 1932	. . 3 ⊢ (∃𝑏[𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
5	2, 4	syl 17	. 2 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
6		nfcv 2902	. . . 4 ⊢ Ⅎ𝑎𝐶
7		nfsbc1v 3797	. . . 4 ⊢ Ⅎ𝑎[𝐴 / 𝑎]𝜑
8	6, 7	nfsbcw 3799	. . 3 ⊢ Ⅎ𝑎[𝐶 / 𝑏][𝐴 / 𝑎]𝜑
9		sbccomieg.1	. . . 4 ⊢ (𝑎 = 𝐴 → 𝐵 = 𝐶)
10		sbceq1a 3788	. . . 4 ⊢ (𝑎 = 𝐴 → (𝜑 ↔ [𝐴 / 𝑎]𝜑))
11	9, 10	sbceqbid 3784	. . 3 ⊢ (𝑎 = 𝐴 → ([𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
12	8, 11	sbciegf 3816	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
13	1, 5, 12	pm5.21nii 378	1 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑)

Syntax hints:   → wi 4   ↔ wb 205   = wceq 1540  ∃wex 1780   ∈ wcel 2105  Vcvv 3473  [wsbc 3777

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2153  ax-12 2170  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-3an 1088  df-tru 1543  df-ex 1781  df-nf 1785  df-sb 2067  df-clab 2709  df-cleq 2723  df-clel 2809  df-nfc 2884  df-ral 3061  df-rex 3070  df-v 3475  df-sbc 3778

This theorem is referenced by:  2rexfrabdioph  41837  3rexfrabdioph  41838  4rexfrabdioph  41839  6rexfrabdioph  41840  7rexfrabdioph  41841