Some simplification rules

Modus ponens: For any two formulas $F$ and $G$, if $F\Rightarrow G$ and $F$ are both known to be true then you may deduce (by modus ponens) that $G$ is also true. More precisely, under any interpretation in which $F\Rightarrow G$ is true and $F$ is true, $G$ is also true.

And-elimination: For any two formulas $F$ and $G$, if $F \wedge G$ is known to be true, then $F$ is true. [This deduction rule is valid because $F \wedge G \Rightarrow F$ is a tautology (do a truth table check if you need to!).]

Or-introduction: For any two formulas $F$ and $G$, if $F$ is known to be true, then $F \vee G$ is true. [This deduction rule is valid because $F \Rightarrow (F \vee G)$ is a tautology (do a truth table check if you need to!).]

  1. Here is a proof that gives the "facts" from each deduction step, but not justifications. Your job is to add for each step the justification, either "given" or the deduction rule (or simplifcation rule) that justify each step. If the deduction step is based on a rule we've given a name to, you can use the name. Otherwise, write down the tautology that justifies the step. 
    Given x => y, ~x => z, and ~z, prove y

1: x => y     ______________________________

2: ~x => z    ______________________________

3: ~z => ~~x  ______________________________

4: ~z         ______________________________

5: ~~x        ______________________________

6: x          ______________________________

7: y	      ______________________________
  2. Here is a puported proof (it's actually invalid!) that gives just the "facts" from each deduction step, but not the justifications. Identify the erroneous step, i.e. the step that is not a valid deduction. This should correspond to a supposed "deduction step" that is not actually a tautology! 
    Given y => z, x and x=>(y|w), prove z

1: x => (y|w)  ______________________________

2: x           ______________________________

3: y|w         ______________________________

4: y           ______________________________

5: y => z      ______________________________

6: z           ______________________________
  3. Given: $x \Rightarrow z$, $w \Rightarrow x \wedge y$, and $w$, Prove $z$.   [Follow the same format as used above!] 

 	EXAMPLE PROBLEM ABOUT LOGICAL ARGUMENT
	Suppose traffic law in a town states:

	LAW: if 20 more more mph over speed limit, then fine must be greater than or equal to $150
	Consider the following two arguments (each is about a different scenario):
	ARG1: "My fine was less than $150, therefore I must not have been 20 or more mph over the limit!"
	ARG2: "Your fine was greater than or equal to $150, therefore you were speeding by 20 or more mph!"
	One of these arguments is logically sound, the other is flawed.  Which?  Why?  How to convince
	others that one is logically sound and the other is not?
Model this argument in propositional logic. Try to justify each step. You will either verify that the argument is sound, or find a step that you cannot justify, which means the argument is not sound.