Math 2090.03 N Assignment 3
due March 5th, 1999

INSTRUCTIONS: This assignment is due March 5th 1999, in class. Groups may hand in a single assignment paper. The maximum group size is 4.

The assignments must have the name and York number of all group members listed at the top of the paper.

Any axiom may be used. Lower numbered theorems may be used in the proof of a higher numbered theorem.


  1. Give a full dress proof that tex2html_wrap_inline44 (tex2html_wrap_inline46 3.76 d)) using proof technique 4.1.

    SOLUTION

    tex2html_wrap_inline48

    Hence by derived inference rule for tex2html_wrap_inline50 number 2, tex2html_wrap_inline44.

  2. Give a full dress proof of modus ponens (tex2html_wrap_inline46 3.77), i.e. tex2html_wrap_inline56 using the Deduction Theorem (4.4).

    SOLUTION

    Assume tex2html_wrap_inline58.

    tex2html_wrap_inline60

    Hence by TRANSITIVITY 3 times and EQUANIMITY tex2html_wrap_inline62. Hence by the Deduction Theorem it follows that tex2html_wrap_inline56.

  3. Give a full dress proof that tex2html_wrap_inline66 using Proof by Cases (4.5). You are allowed to use theorems upto and including Theorem 3.40.

    SOLUTION

    tex2html_wrap_inline68

    gives

    tex2html_wrap_inline70

    tex2html_wrap_inline72 Subs tex2html_wrap_inline74. TRANSITIVITY twice and EQUANIMITY gives that tex2html_wrap_inline76.

    Now tex2html_wrap_inline78 gives

    tex2html_wrap_inline80

    tex2html_wrap_inline82 Subs p:= false. TRANSITIVITY three times and EQUANIMITY gives that tex2html_wrap_inline86.

    Hence by Proof by Cases it follows that tex2html_wrap_inline66.


Stephanie van Willigenburg