Theorem sb5ALTVD 44250

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 2259, 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 43862 is sb5ALTVD 44250 without virtual deductions and was automatically derived from sb5ALTVD 44250.

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 43911	. . . . . 6 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥]𝜑   )
2		equsb1 2484	. . . . . 6 ⊢ [𝑦 / 𝑥]𝑥 = 𝑦
3		sban 2075	. . . . . . 7 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) ↔ ([𝑦 / 𝑥]𝑥 = 𝑦 ∧ [𝑦 / 𝑥]𝜑))
4	3	simplbi2com 502	. . . . . 6 ⊢ ([𝑦 / 𝑥]𝜑 → ([𝑦 / 𝑥]𝑥 = 𝑦 → [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)))
5	1, 2, 4	e10 44031	. . . . 5 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   [𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑)   )
6		spsbe 2077	. . . . 5 ⊢ ([𝑦 / 𝑥](𝑥 = 𝑦 ∧ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
7	5, 6	e1a 43964	. . . 4 ⊢ (   [𝑦 / 𝑥]𝜑   ▶   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   )
8	7	in1 43908	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 → ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
9		hbs1 2257	. . . 4 ⊢ ([𝑦 / 𝑥]𝜑 → ∀𝑥[𝑦 / 𝑥]𝜑)
10		idn2 43950	. . . . . 6 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   (𝑥 = 𝑦 ∧ 𝜑)   )
11		simpr 484	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝜑)
12	10, 11	e2 43968	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝜑   )
13		simpl 482	. . . . . 6 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) → 𝑥 = 𝑦)
14	10, 13	e2 43968	. . . . 5 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   𝑥 = 𝑦   )
15		sbequ1 2232	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 → [𝑦 / 𝑥]𝜑))
16	15	com12 32	. . . . 5 ⊢ (𝜑 → (𝑥 = 𝑦 → [𝑦 / 𝑥]𝜑))
17	12, 14, 16	e22 44008	. . . 4 ⊢ (   ∃𝑥(𝑥 = 𝑦 ∧ 𝜑)   ,   (𝑥 = 𝑦 ∧ 𝜑)   ▶   [𝑦 / 𝑥]𝜑   )
18	9, 17	exinst 43961	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) → [𝑦 / 𝑥]𝜑)
19	8, 18	pm3.2i 470	. 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 406	. 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 44109	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395   = wceq 1533  ∃wex 1773  [wsb 2059

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-10 2129  ax-12 2163  ax-13 2365

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-ex 1774  df-nf 1778  df-sb 2060  df-vd1 43907  df-vd2 43915

This theorem is referenced by: (None)