Theorem sb5ALTVD 45880

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 2310, 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 45493 is sb5ALTVD 45880 without virtual deductions and was automatically derived from sb5ALTVD 45880.

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 = 𝑦 ∧ 𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem sb5ALTVD

Step	Hyp	Ref	Expression
1		idn1 45542	. . . . . 6 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2		equsb1 2521	. . . . . 6 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
3		sban 2117	. . . . . . 7 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
4	3	simplbi2com 508	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)))
5	1, 2, 4	e10 45662	. . . . 5 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)   )
6		spsbe 2119	. . . . 5 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	e1a 45595	. . . 4 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   )
8	7	in1 45539	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
9		hbs1 2308	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10		idn2 45581	. . . . . 6 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   (𝑥 = 𝑦 ∧ 𝜑)   )
11		simpr 490	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜑)
12	10, 11	e2 45599	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝜑   )
13		simpl 488	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦)
14	10, 13	e2 45599	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝑥 = 𝑦   )
15		sbequ1 2284	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
16	15	com12 33	. . . . 5 ⊢ (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
17	12, 14, 16	e22 45639	. . . 4 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   [𝑦 / 𝑥]𝜑   )
18	9, 17	exinst 45592	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
19	8, 18	pm3.2i 476	. 2 ⊢ (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ∧ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑))
20		impbi 211	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) → ((∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))))
21	20	imp 412	. 2 ⊢ ((([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ∧ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)) → ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)))
22	19, 21	e0a 45739	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812  [wsb 2099

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-10 2178  ax-12 2213  ax-13 2402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2100  df-vd1 45538  df-vd2 45546

This theorem is used by: (None)