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Theorem sb5ALTVD 41124
Description: The following User's Proof is a Natural Deduction Sequent Calculus transcription of the Fitch-style Natural Deduction proof of Unit 20 Excercise 3.a., which is sb5 2267, found in the "Answers to Starred Exercises" on page 457 of "Understanding Symbolic Logic", Fifth Edition (2008), by Virginia Klenk. The same proof may also be interpreted as a Virtual Deduction Hilbert-style axiomatic proof. It was completed automatically by the tools program completeusersproof.cmd, which invokes Mel L. O'Cat's mmj2 and Norm Megill's Metamath Proof Assistant. sb5ALT 40736 is sb5ALTVD 41124 without virtual deductions and was automatically derived from sb5ALTVD 41124.
1:: (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2:: [𝑦 / 𝑥]𝑥 = 𝑦
3:1,2: (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦 𝜑)   )
4:3: (   [𝑦 / 𝑥]𝜑   ▶   𝑥(𝑥 = 𝑦𝜑 )   )
5:4: ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑) )
6:: (   𝑥(𝑥 = 𝑦𝜑)   ▶   𝑥(𝑥 = 𝑦𝜑)   )
7:: (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑 )   ▶   (𝑥 = 𝑦𝜑)   )
8:7: (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑 )   ▶   𝜑   )
9:7: (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑 )   ▶   𝑥 = 𝑦   )
10:8,9: (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑 )   ▶   [𝑦 / 𝑥]𝜑   )
101:: ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
11:101,10: (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑 )
12:5,11: (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑 )) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
qed:12: ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑) )
(Contributed by Alan Sare, 21-Apr-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
Assertion
Ref Expression
sb5ALTVD ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
Distinct variable group:   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥,𝑦)

Proof of Theorem sb5ALTVD
StepHypRef Expression
1 idn1 40785 . . . . . 6 (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2 equsb1 2523 . . . . . 6 [𝑦 / 𝑥]𝑥 = 𝑦
3 sban 2077 . . . . . . 7 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
43simplbi2com 503 . . . . . 6 ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦𝜑)))
51, 2, 4e10 40905 . . . . 5 (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦𝜑)   )
6 spsbe 2079 . . . . 5 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
75, 6e1a 40838 . . . 4 (   [𝑦 / 𝑥]𝜑   ▶   𝑥(𝑥 = 𝑦𝜑)   )
87in1 40782 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
9 hbs1 2265 . . . 4 ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10 idn2 40824 . . . . . 6 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   (𝑥 = 𝑦𝜑)   )
11 simpr 485 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝜑)
1210, 11e2 40842 . . . . 5 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   𝜑   )
13 simpl 483 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝑥 = 𝑦)
1410, 13e2 40842 . . . . 5 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   𝑥 = 𝑦   )
15 sbequ1 2239 . . . . . 6 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
1615com12 32 . . . . 5 (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
1712, 14, 16e22 40882 . . . 4 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   [𝑦 / 𝑥]𝜑   )
189, 17exinst 40835 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
198, 18pm3.2i 471 . 2 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
20 impbi 209 . . 3 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) → ((∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))))
2120imp 407 . 2 ((([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑)))
2219, 21e0a 40983 1 ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396   = wceq 1528  wex 1771  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-10 2136  ax-12 2167  ax-13 2381
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-ex 1772  df-nf 1776  df-sb 2061  df-vd1 40781  df-vd2 40789
This theorem is referenced by: (None)
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