Theorem sb5ALTVD 43659

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 43271 is sb5ALTVD 43659 without virtual deductions and was automatically derived from sb5ALTVD 43659.

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 43320	. . . . . 6 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2		equsb1 2490	. . . . . 6 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
3		sban 2083	. . . . . . 7 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
4	3	simplbi2com 503	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)))
5	1, 2, 4	e10 43440	. . . . 5 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)   )
6		spsbe 2085	. . . . 5 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	e1a 43373	. . . 4 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   )
8	7	in1 43317	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
9		hbs1 2265	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10		idn2 43359	. . . . . 6 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   (𝑥 = 𝑦 ∧ 𝜑)   )
11		simpr 485	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜑)
12	10, 11	e2 43377	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝜑   )
13		simpl 483	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦)
14	10, 13	e2 43377	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝑥 = 𝑦   )
15		sbequ1 2240	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
16	15	com12 32	. . . . 5 ⊢ (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
17	12, 14, 16	e22 43417	. . . 4 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   [𝑦 / 𝑥]𝜑   )
18	9, 17	exinst 43370	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
19	8, 18	pm3.2i 471	. 2 ⊢ (([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)) ∧ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑))
20		impbi 207	. . 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 407	. 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 43518	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396   = wceq 1541  ∃wex 1781  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171  ax-13 2371

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-ex 1782  df-nf 1786  df-sb 2068  df-vd1 43316  df-vd2 43324

This theorem is referenced by: (None)