Theorem sbccomieg 43144

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 3752	. 2 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 → 𝐴 ∈ V)
2		spesbc 3834	. . 3 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → ∃𝑏[𝐴 / 𝑎]𝜑)
3		sbcex 3752	. . . 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 2899	. . . 4 ⊢ Ⅎ𝑎𝐶
7		nfsbc1v 3762	. . . 4 ⊢ Ⅎ𝑎[𝐴 / 𝑎]𝜑
8	6, 7	nfsbcw 3764	. . 3 ⊢ Ⅎ𝑎[𝐶 / 𝑏][𝐴 / 𝑎]𝜑
9		sbccomieg.1	. . . 4 ⊢ (𝑎 = 𝐴 → 𝐵 = 𝐶)
10		sbceq1a 3753	. . . 4 ⊢ (𝑎 = 𝐴 → (𝜑 ↔ [𝐴 / 𝑎]𝜑))
11	9, 10	sbceqbid 3749	. . 3 ⊢ (𝑎 = 𝐴 → ([𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
12	8, 11	sbciegf 3781	. 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 206   = wceq 1542  ∃wex 1781   ∈ wcel 2114  Vcvv 3442  [wsbc 3742

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-ex 1782  df-nf 1786  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-nfc 2886  df-ral 3053  df-rex 3063  df-v 3444  df-sbc 3743

This theorem is referenced by:  2rexfrabdioph  43147  3rexfrabdioph  43148  4rexfrabdioph  43149  6rexfrabdioph  43150  7rexfrabdioph  43151