Math 2090.03 N (Winter 98-99) TEST 1A
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 sentence: NOTES (4)
    2. Let p,q,r be boolean variables. Are the following sentences? Give reasons. (6)
      1. tex2html_wrap_inline47 SOLUTION No, missing a ).

      2. tex2html_wrap_inline49 SOLUTION Yes p is a sentence, so tex2html_wrap_inline49 is a sentence.

      3. tex2html_wrap_inline55 SOLUTION No y is not a boolean variable.
  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 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, and EQUANIMITY we have that tex2html_wrap_inline71.

    1. Define soundness. NOTES Is tex2html_wrap_inline43 sound? YES.(3)
    2. Define completeness. NOTES Is tex2html_wrap_inline43 complete? YES. (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,T,F) the sentence evaluates to F. However tex2html_wrap_inline43 is complete, and so the sentence cannot be a theorem.

    1. State Metatheorem 3.7. 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_inline61, and substitution with p:= P gives that tex2html_wrap_inline105. However tex2html_wrap_inline107, and so by Equanimity we have that tex2html_wrap_inline109. Metatheorem 3.7 states that any two theorems are equivalent, and in particular that tex2html_wrap_inline111. Leibniz with E being tex2html_wrap_inline113 gives that tex2html_wrap_inline115, and so by equanimity, tex2html_wrap_inline97.




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Stephanie-van Willigenburg