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Theorem sbcbiVD 45857
Description: Implication form of sbcbii 3795. 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 45521 is sbcbiVD 45857 without virtual deductions and was automatically derived from sbcbiVD 45857.
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 45556 . . . 4 (   𝐴 ∈ 𝐵   ▶   𝐴 ∈ 𝐵   )
2 idn2 45595 . . . . 5 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ∀𝑥(𝜑 ↔ 𝜓)   )
3 spsbc 3752 . . . . 5 (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 ↔ 𝜓) → [𝐴 / 𝑥](𝜑 ↔ 𝜓)))
41, 2, 3e12 45705 . . . 4 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   [𝐴 / 𝑥](𝜑 ↔ 𝜓)   )
5 sbcbig 3790 . . . . 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 45705 . . 3 (   𝐴 ∈ 𝐵   ,   ∀𝑥(𝜑 ↔ 𝜓)   ▶   ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓)   )
87in2 45587 . 2 (   𝐴 ∈ 𝐵   ▶   (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))   )
98in1 45553 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 2145  [wsbc 3739
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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-12 2213  ax-ext 2733
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-sbc 3740  df-vd1 45552  df-vd2 45560
This theorem is used by: (None)
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