Discussion:
Axiom of Duns Scotus
m***@wp.pl
2017-06-11 07:46:55 UTC
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Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.

So, true or false?
William Elliot
2017-06-12 09:04:04 UTC
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Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
Which is it? Dumb or stupid?
Me
2017-06-12 09:10:01 UTC
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Post by m***@wp.pl
So, true or false?
Yes.
g***@gmail.com
2017-06-12 11:38:37 UTC
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Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS

L & L->R -> R

so

R <- L & (not(L) or R)

L = born in march
R = 2+2=5

if L is false then

R <- false & (true or R)
R <- false & true
R <- false

if R is false then

false <- L & (not(L) or false)
false <- L & not(L)
false <- false
true

CONCLUSION: Scotus was wrong!
m***@wp.pl
2017-06-12 13:02:14 UTC
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Post by g***@gmail.com
Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS
L & L->R -> R
Off topic.
Axiom of Duns is - Ex Falso Quodlibet.
You can write it as ~L->(L->R), but it's
not an exact translation.
g***@gmail.com
2017-06-14 02:25:41 UTC
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Post by m***@wp.pl
Post by g***@gmail.com
Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS
L & L->R -> R
Off topic.
Axiom of Duns is - Ex Falso Quodlibet.
You can write it as ~L->(L->R), but it's
not an exact translation.
Okay, I was trying

X
~X

L & L->R -> R

L=X

L & (~L or R) -> R

X & (~X or R) -> R

TRUE & (TRUE or R) -> R

TRUE & TRUE -> R

R

(free variable)
Ross A. Finlayson
2017-06-14 04:27:00 UTC
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Post by g***@gmail.com
Post by m***@wp.pl
Post by g***@gmail.com
Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS
L & L->R -> R
Off topic.
Axiom of Duns is - Ex Falso Quodlibet.
You can write it as ~L->(L->R), but it's
not an exact translation.
Okay, I was trying
X
~X
L & L->R -> R
L=X
L & (~L or R) -> R
X & (~X or R) -> R
TRUE & (TRUE or R) -> R
TRUE & TRUE -> R
R
(free variable)
Ha, free variable!
m***@wp.pl
2017-06-14 06:18:58 UTC
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Post by g***@gmail.com
Post by m***@wp.pl
Post by g***@gmail.com
Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS
L & L->R -> R
Off topic.
Axiom of Duns is - Ex Falso Quodlibet.
You can write it as ~L->(L->R), but it's
not an exact translation.
Okay, I was trying
X
~X
L & L->R -> R
L=X
L & (~L or R) -> R
X & (~X or R) -> R
TRUE & (TRUE or R) -> R
TRUE & TRUE -> R
R
(free variable)
And is R still (2+2=5)?
Ross A. Finlayson
2017-06-15 01:51:59 UTC
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Post by m***@wp.pl
Post by g***@gmail.com
Post by m***@wp.pl
Post by g***@gmail.com
Post by m***@wp.pl
Maciej Wozniak, the author of this thread, was born in march.
True or false?
If Scotus was right, You should have no trouble with the answer.
If You can conclude 2+2=5 from it - it's false
If You can't - it must be true.
So, true or false?
from MODUS PONENS
L & L->R -> R
Off topic.
Axiom of Duns is - Ex Falso Quodlibet.
You can write it as ~L->(L->R), but it's
not an exact translation.
Okay, I was trying
X
~X
L & L->R -> R
L=X
L & (~L or R) -> R
X & (~X or R) -> R
TRUE & (TRUE or R) -> R
TRUE & TRUE -> R
R
(free variable)
And is R still (2+2=5)?
I'd rather see Scotus' axiom as
"the infinite is really infinite".