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7/28/2019 lecture 2- Logic.ppt
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Statement – 8a
7/28/2019 lecture 2- Logic.ppt
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Examples – 8b
7/28/2019 lecture 2- Logic.ppt
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Truth Values of Propositions –
8c
7/28/2019 lecture 2- Logic.ppt
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Statements & Truth Values –
9bT
T
F
F
7/28/2019 lecture 2- Logic.ppt
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Not Propositions – 9c
7/28/2019 lecture 2- Logic.ppt
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Example – 10b
7/28/2019 lecture 2- Logic.ppt
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Example – 10c
7/28/2019 lecture 2- Logic.ppt
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Understanding Statements –
11c
7/28/2019 lecture 2- Logic.ppt
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Example – 11b
7/28/2019 lecture 2- Logic.ppt
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Compound Statement – 12a
7/28/2019 lecture 2- Logic.ppt
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Symbolic Representation –
13a
7/28/2019 lecture 2- Logic.ppt
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Logical Connectives – 14a
7/28/2019 lecture 2- Logic.ppt
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Examples – 14b
7/28/2019 lecture 2- Logic.ppt
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Translating from English to
Symbols – 16
7/28/2019 lecture 2- Logic.ppt
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Translating from English to
Symbols – 17b
7/28/2019 lecture 2- Logic.ppt
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Negation – 19
7/28/2019 lecture 2- Logic.ppt
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Truth Table for ~p – 20
7/28/2019 lecture 2- Logic.ppt
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Conjunction – 21
7/28/2019 lecture 2- Logic.ppt
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Truth Table for p ^ q – 22
7/28/2019 lecture 2- Logic.ppt
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Truth Table for p q – 15
7/28/2019 lecture 2- Logic.ppt
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Truth Table for ~p^q - 2
7/28/2019 lecture 2- Logic.ppt
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Truth Table for ~p^q – 2b
7/28/2019 lecture 2- Logic.ppt
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Truth Table for ~p^q – 2c
7/28/2019 lecture 2- Logic.ppt
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~p ^ (q v~ r) – (2 - 3a)
7/28/2019 lecture 2- Logic.ppt
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~p ^ (q v~ r) – 2 - 3b
7/28/2019 lecture 2- Logic.ppt
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~p ^ (q v~ r) – 2 - 3c
7/28/2019 lecture 2- Logic.ppt
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~p ^ (q v~ r) – 2 - 3d
7/28/2019 lecture 2- Logic.ppt
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Truth Table for ~p (p v~ q) – 2 - 3e v
7/28/2019 lecture 2- Logic.ppt
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Truth Table for (pvq) ^~ (p^q) – 2 - 4a
7/28/2019 lecture 2- Logic.ppt
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Truth Table for (pvq) ^~ (p^q) – 2 -
4c
7/28/2019 lecture 2- Logic.ppt
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Truth Table for (pvq) ^~ (p^q) – 2 -
4e v v
7/28/2019 lecture 2- Logic.ppt
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Truth Table for (pvq) ^~ (p^q) – 2 -
4f
7/28/2019 lecture 2- Logic.ppt
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Symbols for Exclusive OR – 2 - 5a
7/28/2019 lecture 2- Logic.ppt
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52
Conditional: why FF is True
A: YES mathematically and YES by the
truth table.
Here’s a mathematical proof: 1. 0 = 1 (assumption)
2. 1 = 2 (added 1 to both sides)
3. 3 = 6 (multiplied both sides by 3)
7/28/2019 lecture 2- Logic.ppt
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53
Conditional: why FF is True
A: YES mathematically and YES by the
truth table.
Here’s a mathematical proof: 1. 0 = 1 (assumption)
2. 1 = 2 (added 1 to both sides)
3. 3 = 6 (multiplied both sides by 3)
4. 0 = 3 (multiplied no. 1 by 3)