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Naoto Meguro : Amateur. MSC2020: 03B10.
The predicate logic treating the equal sign has defect. You must correct it a little.
Let s(t) be a function symbol.
Let X={∀x(x=x), ∀x(x≠s(x)), ∀x∀y∀z((y=z)→((x=y)↔(x=z)))} be an axiom system.
X has a model in which the object domain is {0,1} and 0=0,1=1,0≠1,1≠0,s(0)=1
and s(1)=0 and is consistent.
Theorem 1. The predicate logic treating the equal sign isn't consistent.
Proof. Let c be an individual symbol. Let a be a free variable. Let p(t) be a predicate symbol.
Define p(t)=(X |- c≠t). p(a)=( X |- c≠a)⇔(X |- ∀x(c≠x))⇒(X |- c≠c)=p(c).
p(a)→p(c) is true. ¬p(c)→(¬p(a)) is true. ¬p(c)→∀x(¬p(x)) should be true.
X |- c=c and X is consistent. So ¬p(c) is true. So ∀x(¬p(x)) should be true.
¬p(s(c)) should be true. But p(s(c)) is true. This is contradiction.♦
The axiom system X"={ p(a)→p(c), ¬p(c), p(s(c))} isn't consistent but you can set up that
the axioms in X" are true. Wrong one is the quantification principle. Let's correct it.
Let L be the set of the theorems of the predicate calculus. Set up that the elements of L
are the logical axioms. If q(a)∈L and a is a free variable,∀xq(x)∈L.
∀x(q(x)→r(x))→(∀yq(y)→(∀zr(z)) ∈L when ∀xq(x) and ∀xr(x) are logical formulas.
Set up that the rule of inference is only deduction principle.
The following theorems are proved then.
Theorem 2. Y∪{q} |- p ⇒ Y |- q→p. (q may include free variables. So X" is consistent.)
Theorem 3. Y∪{q} isn't consistent.⇒ Y |- ¬q.
Theorem 4. Y doesn't include the free variable a and Y |- q(a). ⇒ Y |- ∀xq(x).
Theorem 5. Y doesn't include the individual symbol c and Y |- q(c).⇒ Y |- ∀xq(x).
Theorem 6. Y doesn't include the individual symbol c and is consistent.⇒
Y∪{q(c)→∀q(x)} is consistent.
The completeness theorem is proved by these.
You should correct the predicate logic to evolve the consistent mathematics.
Known theorems won't be lost by the above correction.
1
The predicate logic treating the equal sign has defect. You must correct it a little.
2
Let s(t) be a function symbol.
Let X={∀x(x=x), ∀x(x≠s(x)), ∀x∀y∀z((y=z)→((x=y)↔(x=z)))} be an axiom system.
X has a model in which the object domain is {0,1} and 0=0,1=1,0≠1,1≠0,s(0)=1
and s(1)=0 and is consistent.
Theorem 1. The predicate logic treating the equal sign isn't consistent.
Proof. Let c be an individual symbol. Let a be a free variable. Let p(t) be a predicate symbol.
Define p(t)=(X |- c≠t). p(a)=( X |- c≠a)⇔(X |- ∀x(c≠x))⇒(X |- c≠c)=p(c).
p(a)→p(c) is true. ¬p(c)→(¬p(a)) is true. ¬p(c)→∀x(¬p(x)) should be true.
X |- c=c and X is consistent. So ¬p(c) is true. So ∀x(¬p(x)) should be true.
¬p(s(c)) should be true. But p(s(c)) is true. This is contradiction.♦
The axiom system X"={ p(a)→p(c), ¬p(c), p(s(c))} isn't consistent but you can set up that
the axioms in X" are true. Wrong one is the quantification principle. Let's correct it.
Let L be the set of the theorems of the predicate calculus. Set up that the elements of L
are the logical axioms. If q(a)∈L and a is a free variable,∀xq(x)∈L.
∀x(q(x)→r(x))→(∀yq(y)→(∀zr(z)) ∈L when ∀xq(x) and ∀xr(x) are logical formulas.
Set up that the rule of inference is only deduction principle.
The following theorems are proved then.
Theorem 2. Y∪{q} |- p ⇒ Y |- q→p. (q may include free variables. So X" is consistent.)
Theorem 3. Y∪{q} isn't consistent.⇒ Y |- ¬q.
Theorem 4. Y doesn't include the free variable a and Y |- q(a). ⇒ Y |- ∀xq(x).
Theorem 5. Y doesn't include the individual symbol c and Y |- q(c).⇒ Y |- ∀xq(x).
Theorem 6. Y doesn't include the individual symbol c and is consistent.⇒
Y∪{q(c)→∀q(x)} is consistent.
The completeness theorem is proved by these.
3
You should correct the predicate logic to evolve the consistent mathematics.
Known theorems won't be lost by the above correction.