Various
Trilemmas
The previous section’s “Superman” trilemma, and
associated troika and triad, is a special case of the Equality
Glitch:
A has property P;
B does not have property P;
A equals B
The associated troika is:
Moe: A has property P; B hasn’t property P; A doesn’t equal B
Larry: A has property P; B has property P; A equals B
Curly: A hasn’t property P; B hasn’t property P; A equals B
The associated rule triad
is:
From A
has property P;
And B
does not have property P;
Deduce A
does not equal B
From
A equals B;
And A
has property P;
Deduce B
has property P
From
B does not have property P;
And A
equals B;
Deduce A
does not have property P
Here is the Implication Glitch:
A is true;
A implies B;
B is false.
This trilemma is the voter’s-paradox supported by
this troika:
Moe: A is true, A
implies B, B is true.
Larry: A implies
B, B is false, A is false.
Curly: B is false,
A is true, A does not imply B.
Its rule triad is:
From: A is true; A
implies B; Deduce: B is true.
From: A implies B;
B is false; Deduce: A is false.
From: B is false;
A is true; Deduce: A does not imply B.
The first is Modus Ponens, the second is Modus
Tollens, and the third deserves its own name. I propose “Anti-implication”.
Related to Implication glitch is Equivalence Glitch:
A is true;
A is equivalent to B;
B is false.
Also related to Implication glitch is Modus Ponens Breakdown:
All A are B;
all B are C;
Some A are not C.
I also call this the Barbarism Trilemma because it defies
the classic Aristotelian syllogism, BARBARA. Its rule triad is:
From: All A are B; all B are C; Deduce all A are C.
From: All B are C; some A are not C; Deduce some A are
not B.
From: some A are not C; all A are B; Deduce some B are
not C.
Its troika is:
Moe: All A are B; all B are C; all A are C.
Larry: All B are C; some A are not C; some A are not B.
Curly: some A are not C; all A are B; some B are not C.
Here is the Weak And Glitch:
A is true;
B is true;
A and B is not true
Its troika is:
Moe: A
is true, B is true, (A and B) is true
Larry:
B is true, (A and B) is not true: A is not true
Curly:
(A and B) is not true; A is true; B is not true
Its rule triad is:
From A
is true; B is true; Deduce: (A
and B) is true
From B
is true; (A and B) isn’t true; Deduce: A isn’t true
From
(A and B) isn’t true; A is true; Deduce: B isn’t true
Here is the Strong Or Glitch:
Not-A is true;
Not-B is true;
A or B is true
Its troika is:
Moe: not-A
is true; not-B is true; not-(A or B) is true
Larry:
not-B is true; (A or B) is true; A is true
Curly:
(A or B) is true; not-A is true; B is true
Its rule triad is:
From not-A
is true; not-B is true; Deduce: not-(A or B) is true
From not-B
is true; (A or B) is true; Deduce: A is true
From (A
or B) is true; not-A is true; Deduce: B is true
In
the Reductio Trilemma,
reductio ad absurdum fails:
A
implies B
A
implies not-B
A
It’s
associated with this triad:
From: A implies B; A implies not-B; Deduce: not-A
From: A implies not-B; A; Deduce: A does
not imply B
From: A; A implies B; Deduce: A does not
imply not-B
It’s
associated with this troika:
Moe: not-A
Larry:
A and not-B
Curly:
A and B
Here
is the Reductio Trilemma’s dual; the Failed Dilemma:
A
implies B
Not-A
implies B
Not-B
It’s
associated with this triad:
From:
A implies B; not-A implies B; Deduce: B
From:
not-A implies B; not-B; Deduce: A does not imply B
From:
not-B; A implies B; Deduce: not-A does not imply B
It’s
associated with this troika:
Moe: B
Larry:
not-B and A
Curly:
not-B and not-A
Here is a Linear Loop Trilemma:
A
> B;
B
> C;
C
> A.
For instance:
A crook is better than
a fool;
A fool is better than a
wimp;
A wimp is better than a
crook.
This voter’s paradox is supported by this troika:
Moe:
A crook is better than a fool, and a fool is
better than a wimp.
Larry:
A wimp is better than a crook, and a crook is
better than a fool.
Curly:
A fool is better than a wimp, and a wimp is
better than a crook.
Here is the Constancy Trilemma:
Some X has property P;
Some Y does not have
property P;
Property P is constant.
It is supported by this
troika:
Moe: P(x), not P(y), P
is not constant.
Larry: P is constant,
P(x), P(y)
Curly: not P(y), P is
constant, not P(x)
For instance:
Some men are good;
Some men are not good;
All men are equally good.
It yields this triad:
If some men are good, and
some men are not good,
then not all men are
equally good.
If some men are not good,
and all men are equally good,
then no men are good.
If all men are equally
good, and some men are good,
then all men are
good.
