Working with LMPL Interpretations
Philosophy 12A
June 17, 2010
1 Working with
Given
LMPL Interpretations
Consider the following (
given
) LMPL interpretation
I
1
:
(
I
1
)
F
G
H
I
J
α
+ + - + -
β
- - - + +
γ
+ - - - +
In other words, the interpretation
I
1
has the following features:
D = {
α,β,γ
}
, Ext
(F)
= {
α,γ
}
, Ext
(G)
= {
α
}
,
Ext
(H)
= ∅
(where,
∅
is
the null set
), Ext
(I)
= {
α,β
}
, and Ext
(J)
= {
β,γ
}
.
Question
: What are the
I
1
-truth-values of
‹
–
–
?
‹
∼
Ja
ﬂ
(
∀
x)[Jx
→
(Gx
∨
Fx)]
›
Fc
→
Ic
±
(
∃
x)Gx
→
(
∀
y)(Fy
∨
Gy)
ﬁ
(
∃
x)(Jx
↔
Hx)
–
(
∃
y)(
∀
x)[Gy
(Jx
→
(Ix
∨
Fx))]
Solutions
:
‹
‘
∼
Ja
’ is
true
on
I
1
. This is because ‘
Ja
’ is
false
on
I
1
, since
α
∉
Ext
(J)
.
›
‘
Fc
→
Ic
’ is
false
on
I
1
. This is because its
antecedent
‘
Fc
’ is
true
on
I
1
, since
γ
∈
Ext
(F)
; but its
consequent
‘
Ic
’ is
false
on
I
1
, since
γ
∉
Ext
(I)
.
ﬁ
‘
(
∃
x)(Jx
↔
Hx)
’ is
true
on
I
1
. The
instances
of ‘
(
∃
x)(Jx
↔
Hx)
’ on
I
1
are: (
i
) ‘
Ja
↔
Ha
’, (
ii
)
‘
Jb
↔
Hb
’, and (
iii
) ‘
Jc
↔
Hc
’. Instances (
ii
) and (
iii
) are
false
on
I
1
(
why
?). But, instance (
i
) is
true
on
I
1
, because ‘
Ja
’ and ‘
Ha
’ are
both false
on
I
1
, since
α
∉
Ext
(J)
and
α
∉
Ext
(H)
.
1
ﬂ
‘
(
∀
x)[Jx
→
(Gx
∨
Fx)]
’ is
false
on
I
1
. The
instances
of ‘
(
∀
x)[Jx
→
(Gx
∨
Fx)]
’ on
I
1
are as follows:
(
i
) ‘
Ja
→
(Ga
∨
Fa)
’, (
ii
) ‘
Jb
→
(Gb
∨
Fb)
’, and (
iii
) ‘
Jc
→
(Gc
∨
Fc)
’. Instances (
i
) and (
iii
) are
true
on
I
1
(
why
?). But, instance (
ii
) is
false
on
I
1
. This is because ‘
Jb
’ is
true
on
I
1
, since
β
∈
Ext
(J)
; but ‘
Gb
∨
Fb
’
is
false
on
I
1
, since
β
∉
Ext
(G)
and
β
∉
Ext
(F)
.
2