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Theorem csbeq2gVD 38150
Description: Virtual deduction proof of csbeq2gOLD 37786. The following User's Proof is a Virtual Deduction proof completed automatically by the tools program completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm Megill's Metamath Proof Assistant. csbeq2gOLD 37786 is csbeq2gVD 38150 without virtual deductions and was automatically derived from csbeq2gVD 38150.
 1:: ⊢ (   𝐴 ∈ 𝑉   ▶   𝐴 ∈ 𝑉   ) 2:1: ⊢ (   𝐴 ∈ 𝑉   ▶   (∀𝑥𝐵 = 𝐶 → [𝐴 / 𝑥] 𝐵 = 𝐶)   ) 3:1: ⊢ (   𝐴 ∈ 𝑉   ▶   ([𝐴 / 𝑥]𝐵 = 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶)   ) 4:2,3: ⊢ (   𝐴 ∈ 𝑉   ▶   (∀𝑥𝐵 = 𝐶 → ⦋𝐴 / 𝑥 ⦌𝐵 = ⦋𝐴 / 𝑥⦌𝐶)   ) qed:4: ⊢ (𝐴 ∈ 𝑉 → (∀𝑥𝐵 = 𝐶 → ⦋𝐴 / 𝑥⦌ 𝐵 = ⦋𝐴 / 𝑥⦌𝐶))
(Contributed by Alan Sare, 10-Nov-2012.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
csbeq2gVD (𝐴𝑉 → (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))

Proof of Theorem csbeq2gVD
StepHypRef Expression
1 idn1 37811 . . . 4 (   𝐴𝑉   ▶   𝐴𝑉   )
2 spsbc 3415 . . . 4 (𝐴𝑉 → (∀𝑥 𝐵 = 𝐶[𝐴 / 𝑥]𝐵 = 𝐶))
31, 2e1a 37873 . . 3 (   𝐴𝑉   ▶   (∀𝑥 𝐵 = 𝐶[𝐴 / 𝑥]𝐵 = 𝐶)   )
4 sbceqg 3936 . . . 4 (𝐴𝑉 → ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
51, 4e1a 37873 . . 3 (   𝐴𝑉   ▶   ([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)   )
6 imbi2 337 . . . 4 (([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶) → ((∀𝑥 𝐵 = 𝐶[𝐴 / 𝑥]𝐵 = 𝐶) ↔ (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)))
76biimpcd 238 . . 3 ((∀𝑥 𝐵 = 𝐶[𝐴 / 𝑥]𝐵 = 𝐶) → (([𝐴 / 𝑥]𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶) → (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)))
83, 5, 7e11 37934 . 2 (   𝐴𝑉   ▶   (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶)   )
98in1 37808 1 (𝐴𝑉 → (∀𝑥 𝐵 = 𝐶𝐴 / 𝑥𝐵 = 𝐴 / 𝑥𝐶))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195  ∀wal 1473   = wceq 1475   ∈ wcel 1977  [wsbc 3402  ⦋csb 3499 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590 This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-v 3175  df-sbc 3403  df-csb 3500  df-vd1 37807 This theorem is referenced by: (None)
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