showing that $(lnot lnot Q) to Q$ is not an inutionistic tautology by using an ad hoc 3-valued logic?
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I'm trying to prove that $lnot lnot Q to Q$ is not an intuitionistic tautology by constructing a special finitely-valued logic with strictly more tautologies than intuitonistic logic ... and then showing that that new logic rejects double negation elimination. Is this a valid proof technique?
In particular, there's a three-valued logic that accepts all the axioms of a specific axiomatization of intuitionistic logic (given below). Let's call it $L_3$ . In $L_3$ double negation elimination, i.e. $lnotlnot Qto Q$ is not a tautology.
Can the existence of a logic whose tautologies are a strict superset of intuitionistic logic's tautologies be used to prove that a certain statement is a non-tautology in intuitionistic logic?
By plugging in all possible truth values for the metavariables in the axioms for intuitionistic logic, have I done enough work to show that $L_3$ "agrees" with intuitionistic logic in some sense?
Does it matter that $L_3$ has stray tautologies like triple negation elimination $lnotlnotlnot Q to Q$ ?
Let's write out the axioms of intuitionistic logic.
Then-1
$$ phi to (chi to phi) $$
Then-2
$$ (phi to (chi to psi)) to ((phi to chi) to (phi to psi)) $$
And-1
$$ (phi land chi) to phi $$
And-2
$$ (phi land chi) to chi $$
And-3
$$ phi to (chi to (phi land chi)) $$
Or-1
$$ phi to (phi lor chi) $$
Or-2
$$ chi to (phi lor chi) $$
Or-3
$$ (phi to psi) to ((chi to psi) to ((phi lor chi) to psi)) $$
EFQ
$$ bot to phi $$
The truth tables for our three-valued logic $L_3$ are as follows. The two designated truth values for our tautologies are $T$ and $U$ . Note that implication returns $U$ unless the right argument is $F$, in which case it cycles the three truth values.
AND OR IMP
F U T F U T F U T
(F) F F F (F) F U T (F) U U U
(U) F U U (U) U U T (U) T U U
(T) F U T (T) T T T (T) F U U
As we would hope, $Q to (lnotlnot Q)$ is a tautology but $ (lnotlnot Q) to Q $ is not.
Q→¬¬Q is a tautology, as in intuitionistic logic
T→¬¬T F→¬¬F U→¬¬U
T→ ¬F F→ ¬U U→ ¬T
T→ U F→ T U→ F
U U T
¬¬Q→Q is not a tautology, as in intuitionistic logic
L3 is finitely valued, so we can examine all the cases
¬¬T→T ¬¬F→F ¬¬U→U
¬F→T ¬U→F ¬T→U
U→T T→F F→U
U F U
For reference, here is a Python program I used to check the intuitionistic axioms and a few example tautologies and non-tautologies.
import itertools as it
F = "F"
T = "T"
U = "U"
domain = (F, U, T)
def and_(a, b):
ttab = {
F : {F:F, U:F, T:F},
U : {F:F, U:U, T:U},
T : {F:F, U:U, T:T},
}
return ttab[a][b]
def or_(a, b):
ttab = {
F : {F:F, U:U, T:T},
U : {F:U, U:U, T:T},
T : {F:T, U:T, T:T},
}
return ttab[a][b]
def imp_(a, b):
ttab = {
F : {F:U, U:U, T:U},
U : {F:T, U:U, T:U},
T : {F:F, U:U, T:U},
}
return ttab[a][b]
def is_true(x):
return x == T or x == U
def check_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is not None:
print(func.__name__, counterexample)
assert False
def check_non_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is None:
print(func.__name__, "is unexpectedly a tautology")
assert False
def then1(f, x):
return imp_(f, imp_(x, f))
check_tautology(then1, domain, 2)
def then2(f, x, p):
antecedent = imp_(f, imp_(x, p))
consequent = imp_(imp_(f, x), imp_(f, p))
return imp_(antecedent, consequent)
check_tautology(then2, domain, 3)
def and1(f, x):
return imp_(and_(f, x), f)
check_tautology(and1, domain, 2)
def and2(f, x):
return imp_(and_(f, x), x)
check_tautology(and2, domain, 2)
def and3(f, x):
return imp_(f, imp_(x, and_(f, x)))
check_tautology(and3, domain, 2)
def or1(f, x):
return imp_(f, or_(f, x))
check_tautology(or1, domain, 2)
def or2(f, x):
return imp_(x, or_(f, x))
check_tautology(or2, domain, 2)
def or3(f, x, p):
antecedent1 = imp_(f, p)
antecedent2 = imp_(x, p)
consequent = imp_(or_(f, x), p)
return imp_(antecedent1, imp_(antecedent2, consequent))
check_tautology(or3, domain, 3)
def efq(x):
return imp_(F, x)
check_tautology(efq, domain, 1)
# theorems and nontheorems
def double_negation_introduction(x):
return imp_(x, imp_(imp_(x, F), F))
check_tautology(double_negation_introduction, domain, 1)
def double_negation_elimination_BAD(x):
return imp_(imp_(imp_(x, F), F), x)
check_non_tautology(double_negation_elimination_BAD, domain, 1)
def contrapositive_introduction(x, p):
antecedent = imp_(x, p)
consequent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_tautology(contrapositive_introduction, domain, 2)
def contrapositive_elimination_BAD(x, p):
consequent = imp_(x, p)
antecedent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_non_tautology(contrapositive_elimination_BAD, domain, 2)
proof-verification propositional-calculus intuitionistic-logic
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show 3 more comments
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I'm trying to prove that $lnot lnot Q to Q$ is not an intuitionistic tautology by constructing a special finitely-valued logic with strictly more tautologies than intuitonistic logic ... and then showing that that new logic rejects double negation elimination. Is this a valid proof technique?
In particular, there's a three-valued logic that accepts all the axioms of a specific axiomatization of intuitionistic logic (given below). Let's call it $L_3$ . In $L_3$ double negation elimination, i.e. $lnotlnot Qto Q$ is not a tautology.
Can the existence of a logic whose tautologies are a strict superset of intuitionistic logic's tautologies be used to prove that a certain statement is a non-tautology in intuitionistic logic?
By plugging in all possible truth values for the metavariables in the axioms for intuitionistic logic, have I done enough work to show that $L_3$ "agrees" with intuitionistic logic in some sense?
Does it matter that $L_3$ has stray tautologies like triple negation elimination $lnotlnotlnot Q to Q$ ?
Let's write out the axioms of intuitionistic logic.
Then-1
$$ phi to (chi to phi) $$
Then-2
$$ (phi to (chi to psi)) to ((phi to chi) to (phi to psi)) $$
And-1
$$ (phi land chi) to phi $$
And-2
$$ (phi land chi) to chi $$
And-3
$$ phi to (chi to (phi land chi)) $$
Or-1
$$ phi to (phi lor chi) $$
Or-2
$$ chi to (phi lor chi) $$
Or-3
$$ (phi to psi) to ((chi to psi) to ((phi lor chi) to psi)) $$
EFQ
$$ bot to phi $$
The truth tables for our three-valued logic $L_3$ are as follows. The two designated truth values for our tautologies are $T$ and $U$ . Note that implication returns $U$ unless the right argument is $F$, in which case it cycles the three truth values.
AND OR IMP
F U T F U T F U T
(F) F F F (F) F U T (F) U U U
(U) F U U (U) U U T (U) T U U
(T) F U T (T) T T T (T) F U U
As we would hope, $Q to (lnotlnot Q)$ is a tautology but $ (lnotlnot Q) to Q $ is not.
Q→¬¬Q is a tautology, as in intuitionistic logic
T→¬¬T F→¬¬F U→¬¬U
T→ ¬F F→ ¬U U→ ¬T
T→ U F→ T U→ F
U U T
¬¬Q→Q is not a tautology, as in intuitionistic logic
L3 is finitely valued, so we can examine all the cases
¬¬T→T ¬¬F→F ¬¬U→U
¬F→T ¬U→F ¬T→U
U→T T→F F→U
U F U
For reference, here is a Python program I used to check the intuitionistic axioms and a few example tautologies and non-tautologies.
import itertools as it
F = "F"
T = "T"
U = "U"
domain = (F, U, T)
def and_(a, b):
ttab = {
F : {F:F, U:F, T:F},
U : {F:F, U:U, T:U},
T : {F:F, U:U, T:T},
}
return ttab[a][b]
def or_(a, b):
ttab = {
F : {F:F, U:U, T:T},
U : {F:U, U:U, T:T},
T : {F:T, U:T, T:T},
}
return ttab[a][b]
def imp_(a, b):
ttab = {
F : {F:U, U:U, T:U},
U : {F:T, U:U, T:U},
T : {F:F, U:U, T:U},
}
return ttab[a][b]
def is_true(x):
return x == T or x == U
def check_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is not None:
print(func.__name__, counterexample)
assert False
def check_non_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is None:
print(func.__name__, "is unexpectedly a tautology")
assert False
def then1(f, x):
return imp_(f, imp_(x, f))
check_tautology(then1, domain, 2)
def then2(f, x, p):
antecedent = imp_(f, imp_(x, p))
consequent = imp_(imp_(f, x), imp_(f, p))
return imp_(antecedent, consequent)
check_tautology(then2, domain, 3)
def and1(f, x):
return imp_(and_(f, x), f)
check_tautology(and1, domain, 2)
def and2(f, x):
return imp_(and_(f, x), x)
check_tautology(and2, domain, 2)
def and3(f, x):
return imp_(f, imp_(x, and_(f, x)))
check_tautology(and3, domain, 2)
def or1(f, x):
return imp_(f, or_(f, x))
check_tautology(or1, domain, 2)
def or2(f, x):
return imp_(x, or_(f, x))
check_tautology(or2, domain, 2)
def or3(f, x, p):
antecedent1 = imp_(f, p)
antecedent2 = imp_(x, p)
consequent = imp_(or_(f, x), p)
return imp_(antecedent1, imp_(antecedent2, consequent))
check_tautology(or3, domain, 3)
def efq(x):
return imp_(F, x)
check_tautology(efq, domain, 1)
# theorems and nontheorems
def double_negation_introduction(x):
return imp_(x, imp_(imp_(x, F), F))
check_tautology(double_negation_introduction, domain, 1)
def double_negation_elimination_BAD(x):
return imp_(imp_(imp_(x, F), F), x)
check_non_tautology(double_negation_elimination_BAD, domain, 1)
def contrapositive_introduction(x, p):
antecedent = imp_(x, p)
consequent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_tautology(contrapositive_introduction, domain, 2)
def contrapositive_elimination_BAD(x, p):
consequent = imp_(x, p)
antecedent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_non_tautology(contrapositive_elimination_BAD, domain, 2)
proof-verification propositional-calculus intuitionistic-logic
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2
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You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
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– Hanul Jeon
Dec 6 '18 at 5:39
1
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You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
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– Hanul Jeon
Dec 6 '18 at 5:41
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@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
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– Gregory Nisbet
Dec 6 '18 at 5:48
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One possible method is checking if premises of the inference rules are true then the conclusion is also true.
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– Hanul Jeon
Dec 6 '18 at 5:54
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For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
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– Hanul Jeon
Dec 6 '18 at 5:59
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show 3 more comments
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I'm trying to prove that $lnot lnot Q to Q$ is not an intuitionistic tautology by constructing a special finitely-valued logic with strictly more tautologies than intuitonistic logic ... and then showing that that new logic rejects double negation elimination. Is this a valid proof technique?
In particular, there's a three-valued logic that accepts all the axioms of a specific axiomatization of intuitionistic logic (given below). Let's call it $L_3$ . In $L_3$ double negation elimination, i.e. $lnotlnot Qto Q$ is not a tautology.
Can the existence of a logic whose tautologies are a strict superset of intuitionistic logic's tautologies be used to prove that a certain statement is a non-tautology in intuitionistic logic?
By plugging in all possible truth values for the metavariables in the axioms for intuitionistic logic, have I done enough work to show that $L_3$ "agrees" with intuitionistic logic in some sense?
Does it matter that $L_3$ has stray tautologies like triple negation elimination $lnotlnotlnot Q to Q$ ?
Let's write out the axioms of intuitionistic logic.
Then-1
$$ phi to (chi to phi) $$
Then-2
$$ (phi to (chi to psi)) to ((phi to chi) to (phi to psi)) $$
And-1
$$ (phi land chi) to phi $$
And-2
$$ (phi land chi) to chi $$
And-3
$$ phi to (chi to (phi land chi)) $$
Or-1
$$ phi to (phi lor chi) $$
Or-2
$$ chi to (phi lor chi) $$
Or-3
$$ (phi to psi) to ((chi to psi) to ((phi lor chi) to psi)) $$
EFQ
$$ bot to phi $$
The truth tables for our three-valued logic $L_3$ are as follows. The two designated truth values for our tautologies are $T$ and $U$ . Note that implication returns $U$ unless the right argument is $F$, in which case it cycles the three truth values.
AND OR IMP
F U T F U T F U T
(F) F F F (F) F U T (F) U U U
(U) F U U (U) U U T (U) T U U
(T) F U T (T) T T T (T) F U U
As we would hope, $Q to (lnotlnot Q)$ is a tautology but $ (lnotlnot Q) to Q $ is not.
Q→¬¬Q is a tautology, as in intuitionistic logic
T→¬¬T F→¬¬F U→¬¬U
T→ ¬F F→ ¬U U→ ¬T
T→ U F→ T U→ F
U U T
¬¬Q→Q is not a tautology, as in intuitionistic logic
L3 is finitely valued, so we can examine all the cases
¬¬T→T ¬¬F→F ¬¬U→U
¬F→T ¬U→F ¬T→U
U→T T→F F→U
U F U
For reference, here is a Python program I used to check the intuitionistic axioms and a few example tautologies and non-tautologies.
import itertools as it
F = "F"
T = "T"
U = "U"
domain = (F, U, T)
def and_(a, b):
ttab = {
F : {F:F, U:F, T:F},
U : {F:F, U:U, T:U},
T : {F:F, U:U, T:T},
}
return ttab[a][b]
def or_(a, b):
ttab = {
F : {F:F, U:U, T:T},
U : {F:U, U:U, T:T},
T : {F:T, U:T, T:T},
}
return ttab[a][b]
def imp_(a, b):
ttab = {
F : {F:U, U:U, T:U},
U : {F:T, U:U, T:U},
T : {F:F, U:U, T:U},
}
return ttab[a][b]
def is_true(x):
return x == T or x == U
def check_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is not None:
print(func.__name__, counterexample)
assert False
def check_non_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is None:
print(func.__name__, "is unexpectedly a tautology")
assert False
def then1(f, x):
return imp_(f, imp_(x, f))
check_tautology(then1, domain, 2)
def then2(f, x, p):
antecedent = imp_(f, imp_(x, p))
consequent = imp_(imp_(f, x), imp_(f, p))
return imp_(antecedent, consequent)
check_tautology(then2, domain, 3)
def and1(f, x):
return imp_(and_(f, x), f)
check_tautology(and1, domain, 2)
def and2(f, x):
return imp_(and_(f, x), x)
check_tautology(and2, domain, 2)
def and3(f, x):
return imp_(f, imp_(x, and_(f, x)))
check_tautology(and3, domain, 2)
def or1(f, x):
return imp_(f, or_(f, x))
check_tautology(or1, domain, 2)
def or2(f, x):
return imp_(x, or_(f, x))
check_tautology(or2, domain, 2)
def or3(f, x, p):
antecedent1 = imp_(f, p)
antecedent2 = imp_(x, p)
consequent = imp_(or_(f, x), p)
return imp_(antecedent1, imp_(antecedent2, consequent))
check_tautology(or3, domain, 3)
def efq(x):
return imp_(F, x)
check_tautology(efq, domain, 1)
# theorems and nontheorems
def double_negation_introduction(x):
return imp_(x, imp_(imp_(x, F), F))
check_tautology(double_negation_introduction, domain, 1)
def double_negation_elimination_BAD(x):
return imp_(imp_(imp_(x, F), F), x)
check_non_tautology(double_negation_elimination_BAD, domain, 1)
def contrapositive_introduction(x, p):
antecedent = imp_(x, p)
consequent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_tautology(contrapositive_introduction, domain, 2)
def contrapositive_elimination_BAD(x, p):
consequent = imp_(x, p)
antecedent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_non_tautology(contrapositive_elimination_BAD, domain, 2)
proof-verification propositional-calculus intuitionistic-logic
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I'm trying to prove that $lnot lnot Q to Q$ is not an intuitionistic tautology by constructing a special finitely-valued logic with strictly more tautologies than intuitonistic logic ... and then showing that that new logic rejects double negation elimination. Is this a valid proof technique?
In particular, there's a three-valued logic that accepts all the axioms of a specific axiomatization of intuitionistic logic (given below). Let's call it $L_3$ . In $L_3$ double negation elimination, i.e. $lnotlnot Qto Q$ is not a tautology.
Can the existence of a logic whose tautologies are a strict superset of intuitionistic logic's tautologies be used to prove that a certain statement is a non-tautology in intuitionistic logic?
By plugging in all possible truth values for the metavariables in the axioms for intuitionistic logic, have I done enough work to show that $L_3$ "agrees" with intuitionistic logic in some sense?
Does it matter that $L_3$ has stray tautologies like triple negation elimination $lnotlnotlnot Q to Q$ ?
Let's write out the axioms of intuitionistic logic.
Then-1
$$ phi to (chi to phi) $$
Then-2
$$ (phi to (chi to psi)) to ((phi to chi) to (phi to psi)) $$
And-1
$$ (phi land chi) to phi $$
And-2
$$ (phi land chi) to chi $$
And-3
$$ phi to (chi to (phi land chi)) $$
Or-1
$$ phi to (phi lor chi) $$
Or-2
$$ chi to (phi lor chi) $$
Or-3
$$ (phi to psi) to ((chi to psi) to ((phi lor chi) to psi)) $$
EFQ
$$ bot to phi $$
The truth tables for our three-valued logic $L_3$ are as follows. The two designated truth values for our tautologies are $T$ and $U$ . Note that implication returns $U$ unless the right argument is $F$, in which case it cycles the three truth values.
AND OR IMP
F U T F U T F U T
(F) F F F (F) F U T (F) U U U
(U) F U U (U) U U T (U) T U U
(T) F U T (T) T T T (T) F U U
As we would hope, $Q to (lnotlnot Q)$ is a tautology but $ (lnotlnot Q) to Q $ is not.
Q→¬¬Q is a tautology, as in intuitionistic logic
T→¬¬T F→¬¬F U→¬¬U
T→ ¬F F→ ¬U U→ ¬T
T→ U F→ T U→ F
U U T
¬¬Q→Q is not a tautology, as in intuitionistic logic
L3 is finitely valued, so we can examine all the cases
¬¬T→T ¬¬F→F ¬¬U→U
¬F→T ¬U→F ¬T→U
U→T T→F F→U
U F U
For reference, here is a Python program I used to check the intuitionistic axioms and a few example tautologies and non-tautologies.
import itertools as it
F = "F"
T = "T"
U = "U"
domain = (F, U, T)
def and_(a, b):
ttab = {
F : {F:F, U:F, T:F},
U : {F:F, U:U, T:U},
T : {F:F, U:U, T:T},
}
return ttab[a][b]
def or_(a, b):
ttab = {
F : {F:F, U:U, T:T},
U : {F:U, U:U, T:T},
T : {F:T, U:T, T:T},
}
return ttab[a][b]
def imp_(a, b):
ttab = {
F : {F:U, U:U, T:U},
U : {F:T, U:U, T:U},
T : {F:F, U:U, T:U},
}
return ttab[a][b]
def is_true(x):
return x == T or x == U
def check_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is not None:
print(func.__name__, counterexample)
assert False
def check_non_tautology(func, domain, args):
counterexample = None
for x in it.product(domain, repeat=args):
if not is_true(func(*x)):
counterexample = x
if counterexample is None:
print(func.__name__, "is unexpectedly a tautology")
assert False
def then1(f, x):
return imp_(f, imp_(x, f))
check_tautology(then1, domain, 2)
def then2(f, x, p):
antecedent = imp_(f, imp_(x, p))
consequent = imp_(imp_(f, x), imp_(f, p))
return imp_(antecedent, consequent)
check_tautology(then2, domain, 3)
def and1(f, x):
return imp_(and_(f, x), f)
check_tautology(and1, domain, 2)
def and2(f, x):
return imp_(and_(f, x), x)
check_tautology(and2, domain, 2)
def and3(f, x):
return imp_(f, imp_(x, and_(f, x)))
check_tautology(and3, domain, 2)
def or1(f, x):
return imp_(f, or_(f, x))
check_tautology(or1, domain, 2)
def or2(f, x):
return imp_(x, or_(f, x))
check_tautology(or2, domain, 2)
def or3(f, x, p):
antecedent1 = imp_(f, p)
antecedent2 = imp_(x, p)
consequent = imp_(or_(f, x), p)
return imp_(antecedent1, imp_(antecedent2, consequent))
check_tautology(or3, domain, 3)
def efq(x):
return imp_(F, x)
check_tautology(efq, domain, 1)
# theorems and nontheorems
def double_negation_introduction(x):
return imp_(x, imp_(imp_(x, F), F))
check_tautology(double_negation_introduction, domain, 1)
def double_negation_elimination_BAD(x):
return imp_(imp_(imp_(x, F), F), x)
check_non_tautology(double_negation_elimination_BAD, domain, 1)
def contrapositive_introduction(x, p):
antecedent = imp_(x, p)
consequent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_tautology(contrapositive_introduction, domain, 2)
def contrapositive_elimination_BAD(x, p):
consequent = imp_(x, p)
antecedent = imp_(imp_(p, F), imp_(x, F))
return imp_(antecedent, consequent)
check_non_tautology(contrapositive_elimination_BAD, domain, 2)
proof-verification propositional-calculus intuitionistic-logic
proof-verification propositional-calculus intuitionistic-logic
edited Dec 6 '18 at 17:53
Gregory Nisbet
asked Dec 6 '18 at 5:20
Gregory NisbetGregory Nisbet
568312
568312
2
$begingroup$
You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:39
1
$begingroup$
You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:41
$begingroup$
@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
$endgroup$
– Gregory Nisbet
Dec 6 '18 at 5:48
$begingroup$
One possible method is checking if premises of the inference rules are true then the conclusion is also true.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:54
$begingroup$
For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:59
|
show 3 more comments
2
$begingroup$
You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:39
1
$begingroup$
You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:41
$begingroup$
@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
$endgroup$
– Gregory Nisbet
Dec 6 '18 at 5:48
$begingroup$
One possible method is checking if premises of the inference rules are true then the conclusion is also true.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:54
$begingroup$
For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:59
2
2
$begingroup$
You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:39
$begingroup$
You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:39
1
1
$begingroup$
You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:41
$begingroup$
You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:41
$begingroup$
@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
$endgroup$
– Gregory Nisbet
Dec 6 '18 at 5:48
$begingroup$
@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
$endgroup$
– Gregory Nisbet
Dec 6 '18 at 5:48
$begingroup$
One possible method is checking if premises of the inference rules are true then the conclusion is also true.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:54
$begingroup$
One possible method is checking if premises of the inference rules are true then the conclusion is also true.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:54
$begingroup$
For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:59
$begingroup$
For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:59
|
show 3 more comments
0
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2
$begingroup$
You have to check your 3-valued logic is go well with an inference rule; you did not mention any inference rules of the logic but I do not think your (Hilbert-styled) logic works with no inference rule.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:39
1
$begingroup$
You may interested in Heyting algebras, especially the open set lattice generated by the Sierpinski space.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:41
$begingroup$
@HanulJeon How do I check whether modus ponens works in $L_3$ as an inference rule? $((phi to chi) land phi) to chi$ is always true, but I'm concerned because my definition of $to$ distinguishes between $T$ and $U$ .
$endgroup$
– Gregory Nisbet
Dec 6 '18 at 5:48
$begingroup$
One possible method is checking if premises of the inference rules are true then the conclusion is also true.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:54
$begingroup$
For example, assume that $phi$ and $phito psi$ posses the value T. It happens only if the value of $phi$ and $psi$ are U and F respectively (if I understand correctly), which is impossible. Hence we can say $psi$ has the value T vacuously.
$endgroup$
– Hanul Jeon
Dec 6 '18 at 5:59