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Theorem sb5ALTVD 40082
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 2251, 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 39685 is sb5ALTVD 40082 without virtual deductions and was automatically derived from sb5ALTVD 40082.
 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 = 𝑦 ∧ 𝜑 )   ) 5:4: ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ) 6:: ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ▶   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ) 7:: ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑 )   ▶   (𝑥 = 𝑦 ∧ 𝜑)   ) 8:7: ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑 )   ▶   𝜑   ) 9:7: ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑 )   ▶   𝑥 = 𝑦   ) 10:8,9: ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑 )   ▶   [𝑦 / 𝑥]𝜑   ) 101:: ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑) 11:101,10: ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑 ) 12:5,11: ⊢ (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑 )) ∧ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)) qed:12: ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) )
(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 39734 . . . . . 6 (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2 equsb1 2444 . . . . . 6 [𝑦 / 𝑥]𝑥 = 𝑦
3 sban 2475 . . . . . . 7 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
43simplbi2com 498 . . . . . 6 ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦𝜑)))
51, 2, 4e10 39863 . . . . 5 (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦𝜑)   )
6 spsbe 2015 . . . . 5 ([𝑦 / 𝑥](𝑥 = 𝑦𝜑) → ∃𝑥(𝑥 = 𝑦𝜑))
75, 6e1a 39796 . . . 4 (   [𝑦 / 𝑥]𝜑   ▶   𝑥(𝑥 = 𝑦𝜑)   )
87in1 39731 . . 3 ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑))
9 hbs1 2255 . . . 4 ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10 idn2 39782 . . . . . 6 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   (𝑥 = 𝑦𝜑)   )
11 simpr 479 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝜑)
1210, 11e2 39800 . . . . 5 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   𝜑   )
13 simpl 476 . . . . . 6 ((𝑥 = 𝑦𝜑) → 𝑥 = 𝑦)
1410, 13e2 39800 . . . . 5 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   𝑥 = 𝑦   )
15 sbequ1 2228 . . . . . 6 (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
1615com12 32 . . . . 5 (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
1712, 14, 16e22 39840 . . . 4 (   𝑥(𝑥 = 𝑦𝜑)   ,   (𝑥 = 𝑦𝜑)   ▶   [𝑦 / 𝑥]𝜑   )
189, 17exinst 39793 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)
198, 18pm3.2i 464 . 2 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑))
20 impbi 200 . . 3 (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) → ((∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))))
2120imp 397 . 2 ((([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦𝜑)) ∧ (∃𝑥(𝑥 = 𝑦𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑)))
2219, 21e0a 39941 1 ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦𝜑))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   = wceq 1601  ∃wex 1823  [wsb 2011 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-10 2135  ax-12 2163  ax-13 2334 This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-ex 1824  df-nf 1828  df-sb 2012  df-vd1 39730  df-vd2 39738 This theorem is referenced by: (None)
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