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Theorem sbcbiVD 46118
Description: Implication form of sbcbii 3798. 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 45782 is sbcbiVD 46118 without virtual deductions and was automatically derived from sbcbiVD 46118.
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 45817 . . . 4 (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2 idn2 45856 . . . . 5 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ∀𝑥(𝜑 ↔ 𝜓)   )
3 spsbc 3755 . . . . 5 (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → [𝐴 / 𝑥](𝜑 ↔ 𝜓)))
41, 2, 3e12 45966 . . . 4 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   [𝐴 / 𝑥](𝜑 ↔ 𝜓)   )
5 sbcbig 3793 . . . . 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 45966 . . 3 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)   )
87in2 45848 . 2 (   𝐴 ∈ 𝐵   ▶   (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))   )
98in1 45814 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 1568   ∈ wcel 2147  [wsbc 3742
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-sbc 3743  df-vd1 45813  df-vd2 45821
This theorem is used by: (None)
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