[npl] / trunk / NationalProblemLibrary / SUNYSB / contradiction.pg Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww

Sun Aug 13 01:05:41 2006 UTC (6 years, 9 months ago) by jj
File size: 4701 byte(s)
Fixed typesetting.


    1 ## DESCRIPTION
2 ## Discrete Mathematics
3 ## ENDDESCRIPTION
4
5 ## KEYWORDS('discrete mathematics','logic','proof by contradiction')
6 ## Tagged by cmd6a 8/6/06
7
8 ## DBsubject('Discrete Mathematics')
9 ## DBchapter('Logic')
10 ## DBsection('Reasoning')
11 ## Date('')
12 ## Author('')
13 ## Institution('SUNYSB')
14 ## TitleText1('')
15 ## EditionText1('')
16 ## AuthorText1('')
17 ## Section1('')
18 ## Problem1('')
19
20 DOCUMENT();
22            'PGbasicmacros.pl',
23            'PGchoicemacros.pl',
25 );
26 TEXT(beginproblem());
27 $showPartialCorrectAnswers = 0; 28 29 BEGIN_TEXT 30 For the following$BBOLD proof by contradiction $EBOLD provide the justifications at each step, 31 using the following equivalences and inference rules. Use the following keys: 32 END_TEXT 33 TEXT( 34 begintable(2), 35 row( 'a', 'Idempotent Law'), 36 row( 'b', 'Double Negation'), 37 row( 'c', 'De Morgan~~'s Law'), 38 row( 'd', 'Commutative Properties '), 39 row( 'e', 'Associative Properties '), 40 row( 'f', 'Distributive Properties '), 41 row( 'g', 'Equivalence of Contrapositive '), 42 row( 'h', 'Definition of Implication '), 43 row( 'i', 'Definition of Equivalence ')); 44 BEGIN_TEXT 45 \{ row( 'j', 'Identity Laws $$(p \vee F = p \wedge T = p)$$ ')\} 46 \{ row( 'k', 'Tautology $$(p \vee \neg p = T)$$ ') \} 47 \{ row( 'l', 'Contradiction $$(p \wedge \neg p = F)$$ ') \} 48 END_TEXT 49 TEXT( 50 row( 'm', 'Negation of the goal to prove '), 51 row( 'n', 'Modus Ponens'), 52 row( 'o', 'Modus Tollens'), 53 row( 'p', 'Transitivity of Implication'), 54 row( 'q', 'Conjunctive Simplification'), 55 row( 'r', 'Conjunctive Addition'), 56 row( 's', 'Disjunctive Addition'), 57 endtable() 58 ); 59 60$version = random(1,3,1);
61 if ($version == 1) 62 { 63 64 BEGIN_TEXT 65$PAR
66     We want to prove $$s$$ by a proof by contradiction from the following propositions.
67 END_TEXT
68
69 TEXT( begintable(1));
70 BEGIN_TEXT
71   \{     row( '$$p \rightarrow q$$' ) \}
72   \{     row( '$$r \rightarrow q$$' ) \}
73   \{     row( '$$\neg q$$' ) \}
74   \{     row( '$$\neg(s \wedge T) \rightarrow p$$' ) \}
75 END_TEXT
76 TEXT( endtable() );
77
78 BEGIN_TEXT
79     $PAR 80 $$\neg s$$ by \{ ans_rule(1) \} 81 END_TEXT 82 ANS(str_cmp("m")); 83 84 BEGIN_TEXT 85$PAR
86     $$\neg p$$ by  \{ ans_rule(1) \} between $$p \rightarrow q$$ and $$\neg q$$
87 END_TEXT
88 ANS(str_cmp("o"));
89
90 BEGIN_TEXT
91     $PAR 92 $$s \wedge T$$ by \{ ans_rule(1) \} between $$\neg(s \wedge T) \rightarrow p$$ and $$\neg p$$ previously deduced. 93 END_TEXT 94 ANS(str_cmp("o")); 95 96 BEGIN_TEXT 97$PAR
98     $$s$$ by  \{ ans_rule(1) \} of $$s \wedge T$$
99 END_TEXT
100 ANS(str_cmp("q"));
101
102 BEGIN_TEXT
103     $PAR 104 We have $$s$$ and $$\neg s$$ true, therefore we have a contradiction. 105 END_TEXT 106 } 107 elsif ($version == 2)
108 {
109 BEGIN_TEXT
110     $PAR 111 We want to prove $$d$$ by a proof by contradiction from the following propositions. 112 113 \{ begintable(1) \} 114 \{ row( '$$a \rightarrow b$$' ) \} 115 \{ row( '$$r \rightarrow b$$' ) \} 116 \{ row( '$$\neg b$$' ) \} 117 \{ row( '$$\neg(d \wedge T) \rightarrow a$$' ) \} 118 \{ endtable() \} 119 120$PAR
121     $$\neg d$$ by \{ ans_rule(1) \}
122 END_TEXT
123 ANS(str_cmp("m"));
124
125 BEGIN_TEXT
126     $PAR 127 $$\neg a$$ by \{ ans_rule(1) \} between $$a \rightarrow b$$ and $$\neg b$$ 128 END_TEXT 129 ANS(str_cmp("o")); 130 131 BEGIN_TEXT 132$PAR
133     $$d \wedge T$$ by  \{ ans_rule(1) \} between $$\neg(d \wedge T) \rightarrow a$$ and $$\neg a$$ previously deduced.
134 END_TEXT
135 ANS(str_cmp("o"));
136
137 BEGIN_TEXT
138     $PAR 139 $$d$$ by \{ ans_rule(1) \} of $$d \wedge T$$ 140 END_TEXT 141 ANS(str_cmp("q")); 142 143 BEGIN_TEXT 144$PAR
145     We have $$d$$ and $$\neg d$$ true, therefore we have a contradiction.
146 END_TEXT
147 }
148 else
149 {
150 BEGIN_TEXT
151     $PAR 152 We want to prove $$s$$ by a proof by contradiction from the following propositions. 153 154 \{ begintable(1) \} 155 \{ row( '$$p \rightarrow b$$' ) \} 156 \{ row( '$$r \rightarrow b$$' ) \} 157 \{ row( '$$\neg b$$' ) \} 158 \{ row( '$$\neg(s \wedge T) \rightarrow p$$' ) \} 159 \{ endtable() \} 160 161$PAR
162     $$\neg s$$ by \{ ans_rule(1) \}
163 END_TEXT
164 ANS(str_cmp("m"));
165
166 BEGIN_TEXT
167     $PAR 168 $$\neg p$$ by \{ ans_rule(1) \} between $$p \rightarrow b$$ and $$\neg b$$ 169 END_TEXT 170 ANS(str_cmp("o")); 171 172 BEGIN_TEXT 173$PAR
174     $$s \wedge T$$ by  \{ ans_rule(1) \} between $$\neg(s \wedge T) \rightarrow p$$ and $$\neg p$$ previously deduced.
175 END_TEXT
176 ANS(str_cmp("o"));
177
178 BEGIN_TEXT
179     $PAR 180 $$s$$ by \{ ans_rule(1) \} of $$s \wedge T$$ 181 END_TEXT 182 ANS(str_cmp("q")); 183 184 BEGIN_TEXT 185$PAR
186     We have $$s$$ and $$\neg s$$ true, therefore we have a contradiction.
187 END_TEXT
188 }
189
190 ENDDOCUMENT();
191
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195