2.66 Let K1 be a theory whose language has only = as a predicate letter and no f

Question

2.66 Let K1 be a theory whose language has a predicate letter and no function letters or individual constants. Let its proper axioms be (for all x1)x1 = x1,(for all x1)(for all x2)(x1 = x2 right arrow x2 = x1), and (for all x1)(for all x2)(for all x3)(x1 = x2 right arrow (x2 = x3 right arrow x1 = x3)). Show that K1 is a theory with equality. [Hint: It suffices to prove that X1 = X3 right arrow (X1 = X2 right arrow X3 = X2) and X2 = x3 right arrow (x1 = x2 right arrow x1 = x3). K1 is called the pure first-order theory of equality.

Explanation / Answer

K1 be theory:

Hint: x1=x3 implies (x1=x2 implies x3=x2)

   x2=x3 implies (x1=x2 implies x1=x3)

   let us consider an example x1=x3 f(x)=ax+b => ax1+b=ax3+b => x1=x3

   f(x)=ax+b

   given x1=x2

   f(x1)=f(x2)

   (ax1+b)=(ax2+b)

   ax1+b=ax2+b =>ax1=ax2

   x1=x2

     now,consider x3=x2 f(x)=ax+b

   f(x3)=f(x2)

   ax3+b=ax2+b

   x3=x2

   x1=x3 => (x1=x2 => x3=x2)

   let,consider f(x2)=f(x3)

   f(x)=cx+d

   cx2+d=cx3+d

   x2=x3

   let,f(x)=ax+c

   f(x1)=f(x2)

   ax1+c=ax2+c

   x1=x2

   let,f(x)=ax+b

   x1=x3

   f(x1)=f(x3)

   ax1+b=ax3+b

   x1=x3

   x2=x3 => (x1=x2 => x1=x3)

     

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.