Theorem sbccomieg 43334

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 3754	. 2 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 → 𝐴 ∈ V)
2		spesbc 3835	. . 3 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → ∃𝑏[𝐴 / 𝑎]𝜑)
3		sbcex 3754	. . . 4 ⊢ ([𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
4	3	exlimiv 1949	. . 3 ⊢ (∃𝑏[𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
5	2, 4	syl 17	. 2 ⊢ ([𝐶 / 𝑏][𝐴 / 𝑎]𝜑 → 𝐴 ∈ V)
6		nfcv 2923	. . . 4 ⊢ Ⅎ𝑎𝐶
7		nfsbc1v 3764	. . . 4 ⊢ Ⅎ𝑎[𝐴 / 𝑎]𝜑
8	6, 7	nfsbcw 3766	. . 3 ⊢ Ⅎ𝑎[𝐶 / 𝑏][𝐴 / 𝑎]𝜑
9		sbccomieg.1	. . . 4 ⊢ (𝑎 = 𝐴 → 𝐵 = 𝐶)
10		sbceq1a 3755	. . . 4 ⊢ (𝑎 = 𝐴 → (𝜑 ↔ [𝐴 / 𝑎]𝜑))
11	9, 10	sbceqbid 3751	. . 3 ⊢ (𝑎 = 𝐴 → ([𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
12	8, 11	sbciegf 3782	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑))
13	1, 5, 12	pm5.21nii 380	1 ⊢ ([𝐴 / 𝑎][𝐵 / 𝑏]𝜑 ↔ [𝐶 / 𝑏][𝐴 / 𝑎]𝜑)

Syntax hints:   → wi 4   ↔ wb 208   = wceq 1559  ∃wex 1798   ∈ wcel 2141  Vcvv 3453  [wsbc 3744

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ral 3076  df-rex 3086  df-v 3455  df-sbc 3745

This theorem is referenced by:  2rexfrabdioph  43337  3rexfrabdioph  43338  4rexfrabdioph  43339  6rexfrabdioph  43340  7rexfrabdioph  43341