Math 2090.03 N (Winter 98-99) TEST 1B
Friday February 12th, 1999
SOLUTIONS

SURNAME (BLOCK CAPITALS):

INITIALS:

STUDENT NUMBER:

SIGNATURE:

All questions concern our propositional logic tex2html_wrap_inline43 and material covered in Chapter 3 of the course text. The marks for each question are shown in brackets.


    1. Define theorem (in tex2html_wrap_inline43): NOTES (6)
    2. Let p,q,r be boolean variables. Give reasons why the following are theorems in tex2html_wrap_inline43. (4)
      1. tex2html_wrap_inline51 SOLUTION Axiom 3.3.
      2. tex2html_wrap_inline53 SOLUTION Axiom 3.28, Subs p:=q.
      3. tex2html_wrap_inline57 SOLUTION Axiom 3.57.
  1. Fill in the tex2html_wrap_inline59 to give a full dress proof that tex2html_wrap_inline61 (tex2html_wrap_inline63). All axioms, Metatheorems from Chapter 3, and lower numbered theorems may be used in this proof. (10)

    tex2html_wrap_inline65

    Hence by TRANSITIVITY 3 times, and EQUANIMITY, we have that tex2html_wrap_inline61.

  2. Fill in the tex2html_wrap_inline59 to give a full dress proof that tex2html_wrap_inline71 (tex2html_wrap_inline73). All axioms, Metatheorems from Chapter 3, and lower numbered theorems may be used in this proof. (10)

    tex2html_wrap_inline75

    Hence by TRANSITIVITY 3 times tex2html_wrap_inline71.

    1. Define completeness. NOTES tex2html_wrap_inline43 is not complete. (Yes/No) NO.(3)
    2. Define soundness. NOTES tex2html_wrap_inline43 is not sound. (Yes/No) NO.(3)
    3. Let p,q,r be boolean variables. Explain why tex2html_wrap_inline85 is not a theorem in tex2html_wrap_inline43.(4)

      SOLUTION In state (p,q,r)=(T,F,F) the sentence evaluates to F. However tex2html_wrap_inline43 is complete, and so the sentence cannot be a theorem.

    1. State Metatheorem ``Redundant True''. NOTES (2)
    2. Prove that if tex2html_wrap_inline95 then tex2html_wrap_inline97. You may use anything from Chapter 3 except those facts concerning soundness and completeness. (8)

      SOLUTION Assume tex2html_wrap_inline95. However tex2html_wrap_inline101, and substitution with p:= P gives that tex2html_wrap_inline105. However by Metatheorem ``Redundant True'' and Equanimity we have that tex2html_wrap_inline107. Metatheorem 3.7 states that any two theorems are equivalent, and in particular that tex2html_wrap_inline109. Leibniz with E being tex2html_wrap_inline111 gives that tex2html_wrap_inline113, and so by equanimity, tex2html_wrap_inline97.




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Stephanie-van Willigenburg