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Theorem sbcbiVD 45902
Description: Implication form of sbcbii 3794. 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. sbcbi 45566 is sbcbiVD 45902 without virtual deductions and was automatically derived from sbcbiVD 45902.
1:: (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2:: (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)    ▶   ∀𝑥(𝜑 ↔ 𝜓)   )
3:1,2: (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)    ▶   [𝐴 / 𝑥](𝜑 ↔ 𝜓)   )
4:1,3: (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)    ▶   ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)   )
5:4: (   𝐴 ∈ 𝐵   ▶   (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))   )
qed:5: (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)))
(Contributed by Alan Sare, 18-Mar-2012.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sbcbiVD (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)))

Proof of Theorem sbcbiVD
StepHypRef Expression
1 idn1 45601 . . . 4 (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2 idn2 45640 . . . . 5 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ∀𝑥(𝜑 ↔ 𝜓)   )
3 spsbc 3751 . . . . 5 (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → [𝐴 / 𝑥](𝜑 ↔ 𝜓)))
41, 2, 3e12 45750 . . . 4 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   [𝐴 / 𝑥](𝜑 ↔ 𝜓)   )
5 sbcbig 3789 . . . . 5 (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)))
65biimpd 232 . . . 4 (𝐴 ∈ 𝐵 → ([𝐴 / 𝑥](𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)))
71, 4, 6e12 45750 . . 3 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)   )
87in2 45632 . 2 (   𝐴 ∈ 𝐵   ▶   (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))   )
98in1 45598 1 (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   ∈ wcel 2144  [wsbc 3738
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-12 2212  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739  df-vd1 45597  df-vd2 45605
This theorem is used by: (None)
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