ΑΝΑΛΥΣΗ ΕΝΟΣ ΑΠΟΣΠΑΣΜΑΤΟΣ ΑΠΟ ΤΟ «DEMONSTRATION OF THE BEING AND...

8/2/2019 DEMONSTRATION OF THE BEING AND ATTRIBUTES OF GOD, DR. SA

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XIII

DEMONSTRATION OF THE BEING

AND ATTRIBUTES OF GOD, DR. SAMUEL CLARKE

ETHICA ORDINE GEOMETRICO

DEMONSTRATA SPINOZA.

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3. Demonstration of the Being and Attributes of God, ,

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3/33

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4. Ethics Benedict Spinoza , , ,

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. Dr. Samuel Clarke ,

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CLARKE.

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4/33

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Dr. Clarke


5/33

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x = .

y = .

z = .

p = .

q = .

, , x = " ,

x (XI.

7), x = 1, x = 0, .

:-

1

x = 1

2 x = v[y(1x) + z(1y)]

3

x = v[p(1q) + q(1p)]

4

p = vy

5

q = v(1z)

v,


6/33

1x = 0 (1)

x = [yz + (1y) (1z)] = 0 (2)

x = [pq + (1p) (1q)] = 0 (3)

p(qy) = 0 (4)

qz = 0 (5)

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y, z, p, q, , ,

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y,

z, .

.

x (1), (2), (3),

(2) (3) x (1). ,

yz +(1y)(1z)=0 (6)

pq +(1p)(1q)=0 (7)

p (4) (7), ,

p(1y) + pq + (1p)(1q) = 0 (8)

,

(1y)(1q)=0. (9)

q (5) (9),

qz + (1y)(1q) = 0

x(1y) = 0. (10)

(6) (10), , , ,

yz + 1y = 0.


7/33

z,

1y = 0, , y = 1. (11)

y,

z = 0. (12)

(11)

.

(12)

.

(6), (7), (4), (5),

p q . y (4)

(6),

p(1y) + yz +(1y)(1z) = (0)

(p + 1z)z = 0, or, pz = 0. (13)

z (5) (13),

qz + pz = 0

,

0 = 0.

(7), q, 0 = 0

, p.

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. , p,

0 = 0, q, ,

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(7) p q,


8/33

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(7),

p(1q) + q(1p) = 1, ,

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Dr. Clarke,

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II.

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9/33

x = .y = .

z = .

p = .

q = .

, :

1

x = 1

2

x = v{y(lz) + z(1y)}

3

z = v{p(lq) + (1p)q}

4 p = 0

5

q = 0,

, v,

lx = 0 (1)

x{yz + (1y)(1z)} = 0 (2)

z{pq + (1p)(1q)} = 0 (3)

p = 0 (4)

q = 0. (5)

x, p, q, y, z = 0.

x, p, q, z,

y=1.

:

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2.

Dr. Clarke . ,

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10/33

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2

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3

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x = .

y = .

z = .

w = .

x(1y)(1z) + y(1x)(1z) + z(1x)(1y) = l, (1)


11/33

x = 0, (2)

w=v(1y), (3)

v,

wy = 0. (4)

, , 0 1,

. x

y(1z) + z(1y) = 1 (5)

, yz +(1y)(1z) = 0. (6)

(4) (6), y,

w(1z) = 0,

w = vz

,-.

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9. Dr. Samuel Clarke ,

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( Sir Isaac Newton

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x = .

y = .

t = .

z = ,

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w = .

v = .

,

q :

x = q{y(1t) + (1y)t}.

tz = q(1w).

y = qv.

v = q(1x).

x = qz.

w = 1.


13/33

, q, ,

.:

x{yt +(1 y)(1 t)} = 0. (1)

tzw = 0. (2)

y(1 v) = 0. (3)

vx = 0. (4)

x(1 z) = 0. (5)

1 w = 0. (6)

w, v, z, y, t,

x = 0,

, .

Dr. Clarke.

( v, z, y, t, x, w = 1),

0 = 0.

.

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tz = 0,

z = q(1 t) .

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10. Dr. Clarke ,

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IV. .

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VI.

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4

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x = .

y = .

z = .

w = .

t = .


15/33

, , ,

xy(1 z) = 0. (1)

x(1w) = 0. (2)

w(1t) = 0. (3)

tz = 0. (4)

t,w, z,

xy = 0,

y = 0/0 (1x);

,- .

VII. .

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x = .

y = .

z = .

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x(1y) = 0. (1)


16/33

y(1z) = 0. (2)

zx = 0. (3)

y z,

x = 0.

,.

11. , Bishop Butler, Demonstration,

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Dr. Clarke,

Butler. Butler :

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Dr. Clarke .

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18/33

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x = ( )

1x = .

y = .

p = .

q = .

r = .

s = .

x = vy.

1x = v {p (1q) (1r) + q (1p) (1r) + r (1p) (1q)}.

p = vy.

q = vs (1s) = 0.

r (1q) (1p) = 0.

, , q = 0, , q

,

x = vy, 1x = v {p (1r) + r (1p)},

p = vy, r (1p) = 0,

v, ,

x (1y)=0, (1)

(1x) {pr +(1p)(1r)} = 0, (2)

p(1y) = 0, (3)


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r (1p) = 0. (4)

y,

Dr. Clarke. , x (1) (2),

(1y){pr + (1p)(1r)} = 0. (5)

, r (4) (5),

r(lp) + (1y){pr + (1p)(1r)} = 0,

{1p + (1y)p} x (1y)(1p) = 0;

(1y)(1p) = 0. (6)

, p (3) (6),

1y = 0,

y = 1 ,

, .

.

y (1), (2), (3),

(4), , ,

(1x){pr +(1p)(1r)} = 0, r(1p) = 0. (7)

x,

r(1p) = 0, r = vp, ,

, .

, , r,

(1x)(1p) = 0,

, 1x = vp,

,,

.

, p,

(1x) r = 0,

,

1x = r(1r),


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r = vx,

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, (2) (4)

px = 0,

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(1x) {pr + (1p)(1r)} + r(1p) + px = 0. (8)

p,

r = 0,

, .

r (8),

(1x)(1p) + px = 0, or, 1x = p

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w = .

x = .

y = .

z = .

p = .

q = .

r = .

t = .

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:

w = v {x(1y)(1q) + y(1x)(1z) + z(1x)(1y)}.

x = v(1t).


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y = v {p(1q)(1r) + q(1p)(1r) + r(1p)(1q)}.

p(lq) + q(lp) = v(1t).

t = vw.

, v,

w {1x(1y)(1z)y(1x)(1z)z(1x)(1y)} = 0, (1)

xt = 0, (2)

y {1p(1q)(1r)q(1p)(1r)r(1p)(1q)} = 0, (3)

{p(1q) + q(1p)} t = 0, (4)

t(1w) = 0. (5)

, ,

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t z r.

w, x, y, p, q . ,

.

w (1) (5),

t{1x(1y)(1z)y(1x)(1z)z(1x)(1y)} = 0. (6)

p (3) (4)

yqr + yqt + yt(1r)(1q) = 0. (7)

q

yt(1r) = 0. (8)

x (2) (6)

t{yz + (1y)(1z)} = 0. (9)

y (8) (9)

t(1z)(1r) =0.

t, z, r.

t = 0/{(1z)(1r)}

= (0/0)zr + (0/0)z(1r)+(0/0)(1z)r+0(1z)(1r)

= (0/0)z + (0/0)(1z)r


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, , , ,

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. (8)

t=0/y(1r) = (0/0)yr + (0/0) (1y)r + 0/0(1y)(1r)=(0/0)yr + (0/0)(1y)

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1.

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z, p, q, 0 = 0.

2. ,

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x, t, p

xt + pt = 0;

t = 0/(p+x)=(0/0)=(1p)(1x).

, , .

3.,

.

pqy = 0,

y = 0/pq = (0/0)p(1q)+(0/0)(1p).

,

, .

4.

;


25/33

y(1z) = s, , ,

ts(1r) = 0

s = 0/t(1r )=(0/0)tr+0/0(1t).

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1t = yz + (0/0)y(1z) + (0/0)(1y)z + (1y)(1z).

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15. , , , , ,

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1. (w) (x), (y), (z).2. (x) (t).3. (y) (p), (q), (r).


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4. (p) (t).5. (q) (t).6. (z) (u).7. (r) (u).

8. .

,

v, :

w(1x)(1y)(1z) = 0, (1)

xt = 0, (2)

y(1p)(1q)(1r) = 0, (3)

pt = 0, (4)

qt = 0, (5)

z(1u) = 0, (6)

r(1u) = 0, (7)

t(1w) = 0. (8)

V(1V) = 0.

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16. . . Spinoza.

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1. (causa sui), , .

2. (in suo genere finita) ..

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28/33

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4. .5. .

6. . (Idea vera debet cum suoideato convenire.)

7. . , ,

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V. .

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Ethics ,

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Spinoza.


29/33

17. , , , Spinoza ,

1

, x,

, x' , ,

x + x' = 1; (. I.)

, x = 1 - x'.

2

, y,

, y'

y =1 - y'. (. II.)

3

, z, , z'

z =1 - z'. (. III.,V.)

4

, f, , f';

f =1 - f'. (. VII.)

5

, e,

, e'

e =1 - e'. (. I. . VII.)

,

, . . III.,

,

z = y.

, . IV., Spinoza,

y = e.

, . VII.,

f = e.

, . V

z' = x', ,

z = x.

,

:


30/33

x = y = z = f = e = 1 - x' = 1 - y' = 1 - f' = 1 - z' = 1 - e'.

, Spinoza ,

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z = 1 - e',

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z = e,

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. Spinoza

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: Quum finitum esse revera sit ex parte negatio, et infinitum absoluta

affirmatio existentiae alicujus naturae, sequitur ergo ex sola Prop. VII. omnem substantiam

debere esse infinitam.

Clarke, Butler (. 11),

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, a priori,

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. Spinoza

: Mens humana adaequatum habet cogni-tionem aeternae et infinitae

essentiae Dei (. XLVII., 2). . XXXIV., 2

Omnis idea quae in nobis est absoluta sive adaequata et perfecta, vera est

VI., 1, Idea vera debet cum suo ideato convenire. ,

: De natura rationis est res sub quadam aeternitatis

specie percipere (. XLIV., 2). ,

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Transcript of ΑΝΑΛΥΣΗ ΕΝΟΣ ΑΠΟΣΠΑΣΜΑΤΟΣ ΑΠΟ ΤΟ «DEMONSTRATION OF THE BEING AND...