sb5ALTVD

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

		Ref	Expression
	Assertion	sb5ALTVD	$⊢ [y/ x] φ \leftrightarrow \exists x (x = y \land φ)$

Proof

Step	Hyp	Ref	Expression
1		idn1	$⊢ ([y/ x] φ \to [y/ x] φ)$
2		equsb1	$⊢ [y/ x] x = y$
3		sban	$⊢ [y/ x] (x = y \land φ) \leftrightarrow ([y/ x] x = y \land [y/ x] φ)$
4	3	simplbi2com	$⊢ [y/ x] φ \to ([y/ x] x = y \to [y/ x] (x = y \land φ))$
5	1 2 4	e10	$⊢ ([y/ x] φ \to [y/ x] (x = y \land φ))$
6		spsbe	$⊢ [y/ x] (x = y \land φ) \to \exists x (x = y \land φ)$
7	5 6	e1a	$⊢ ([y/ x] φ \to \exists x (x = y \land φ))$
8	7	in1	$⊢ [y/ x] φ \to \exists x (x = y \land φ)$
9		hbs1	$⊢ [y/ x] φ \to \forall x [y/ x] φ$
10		idn2	$⊢ (\exists x (x = y \land φ), (x = y \land φ) \to (x = y \land φ))$
11		simpr	$⊢ (x = y \land φ) \to φ$
12	10 11	e2	$⊢ (\exists x (x = y \land φ), (x = y \land φ) \to φ)$
13		simpl	$⊢ (x = y \land φ) \to x = y$
14	10 13	e2	$⊢ (\exists x (x = y \land φ), (x = y \land φ) \to x = y)$
15		sbequ1	$⊢ x = y \to (φ \to [y/ x] φ)$
16	15	com12	$⊢ φ \to (x = y \to [y/ x] φ)$
17	12 14 16	e22	$⊢ (\exists x (x = y \land φ), (x = y \land φ) \to [y/ x] φ)$
18	9 17	exinst	$⊢ \exists x (x = y \land φ) \to [y/ x] φ$
19	8 18	pm3.2i	$⊢ (([y/ x] φ \to \exists x (x = y \land φ)) \land (\exists x (x = y \land φ) \to [y/ x] φ))$
20		impbi	$⊢ ([y/ x] φ \to \exists x (x = y \land φ)) \to ((\exists x (x = y \land φ) \to [y/ x] φ) \to ([y/ x] φ \leftrightarrow \exists x (x = y \land φ)))$
21	20	imp	$⊢ (([y/ x] φ \to \exists x (x = y \land φ)) \land (\exists x (x = y \land φ) \to [y/ x] φ)) \to ([y/ x] φ \leftrightarrow \exists x (x = y \land φ))$
22	19 21	e0a	$⊢ [y/ x] φ \leftrightarrow \exists x (x = y \land φ)$

1::	\|- (. [ y / x ] ph ->. [ y / x ] ph ).
2::	\|- [ y / x ] x = y
3:1,2:	\|- (. [ y / x ] ph ->. [ y / x ] ( x = y /\ ph ) ).
4:3:	\|- (. [ y / x ] ph ->. E. x ( x = y /\ ph ) ).
5:4:	\|- ( [ y / x ] ph -> E. x ( x = y /\ ph ) )
6::	\|- (. E. x ( x = y /\ ph ) ->. E. x ( x = y /\ ph ) ).
7::	\|- (. E. x ( x = y /\ ph ) ,. ( x = y /\ ph ) ->. ( x = y /\ ph ) ).
8:7:	\|- (. E. x ( x = y /\ ph ) ,. ( x = y /\ ph ) ->. ph ).
9:7:	\|- (. E. x ( x = y /\ ph ) ,. ( x = y /\ ph ) ->. x = y ).
10:8,9:	\|- (. E. x ( x = y /\ ph ) ,. ( x = y /\ ph ) ->. [ y / x ] ph ).
101::	\|- ( [ y / x ] ph -> A. x [ y / x ] ph )
11:101,10:	\|- ( E. x ( x = y /\ ph ) -> [ y / x ] ph )
12:5,11:	\|- ( ( [ y / x ] ph -> E. x ( x = y /\ ph ) ) /\ ( E. x ( x = y /\ ph ) -> [ y / x ] ph ) )
qed:12:	\|- ( [ y / x ] ph <-> E. x ( x = y /\ ph ) )

Metamath Proof Explorer

Theorem sb5ALTVD

Proof