Mixing
![Find $alpha$,$beta$ if $lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x -beta] = 0$](https://lh6.googleusercontent.com/-SKcyU5lCbpE/AAAAAAAAAAI/AAAAAAAAAAA/AAN31DXcuzLcXi9U7ekHEa1DnjucMea5-g/s72-c-mo/photo.jpg?sz=32)
5 Lined Rules of Inference Question
$begingroup$
Use the rules of inference together with basic
logical equivalences to show that the following argument is valid. Name the
rule you use at each step.
w ∨ ¬z → r
s ∨ ¬w
¬t
z → t
¬z ∧ r → ¬s
—————–
∴ ¬w
I'm really not sure how to work through this problem, I've never worked on a 5 line inference question so I'm not sure how to grasp this.
propositional-calculus
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
$endgroup$
add a comment |
$begingroup$
Use the rules of inference together with basic
logical equivalences to show that the following argument is valid. Name the
rule you use at each step.
w ∨ ¬z → r
s ∨ ¬w
¬t
z → t
¬z ∧ r → ¬s
—————–
∴ ¬w
I'm really not sure how to work through this problem, I've never worked on a 5 line inference question so I'm not sure how to grasp this.
propositional-calculus
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
$endgroup$
add a comment |
$begingroup$
Use the rules of inference together with basic
logical equivalences to show that the following argument is valid. Name the
rule you use at each step.
w ∨ ¬z → r
s ∨ ¬w
¬t
z → t
¬z ∧ r → ¬s
—————–
∴ ¬w
I'm really not sure how to work through this problem, I've never worked on a 5 line inference question so I'm not sure how to grasp this.
propositional-calculus
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
$endgroup$
Use the rules of inference together with basic
logical equivalences to show that the following argument is valid. Name the
rule you use at each step.
w ∨ ¬z → r
s ∨ ¬w
¬t
z → t
¬z ∧ r → ¬s
—————–
∴ ¬w
I'm really not sure how to work through this problem, I've never worked on a 5 line inference question so I'm not sure how to grasp this.
propositional-calculus
propositional-calculus
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
asked Jan 25 at 10:45
Joshua FreemanJoshua Freeman
33
33
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
With $lnot t$ and $z to t$ derive $lnot z$ [with Modus tollens].
With $lnot z$ and $w ∨ ¬z → r$ derive $r$ [with Addition and Modus ponens].
With $r$ and $lnot z$ and $¬z ∧ r → ¬s$ derive $lnot s$ [with Conjunction and Modus ponens].
With $lnot s$ and $s ∨ ¬w$ derive $¬w$ [with Disjunctive syllogism].
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Hint
With $lnot t$ and $z to t$ derive $lnot z$ [with Modus tollens].
With $lnot z$ and $w ∨ ¬z → r$ derive $r$ [with Addition and Modus ponens].
With $r$ and $lnot z$ and $¬z ∧ r → ¬s$ derive $lnot s$ [with Conjunction and Modus ponens].
With $lnot s$ and $s ∨ ¬w$ derive $¬w$ [with Disjunctive syllogism].
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
add a comment |
$begingroup$
Hint
With $lnot t$ and $z to t$ derive $lnot z$ [with Modus tollens].
With $lnot z$ and $w ∨ ¬z → r$ derive $r$ [with Addition and Modus ponens].
With $r$ and $lnot z$ and $¬z ∧ r → ¬s$ derive $lnot s$ [with Conjunction and Modus ponens].
With $lnot s$ and $s ∨ ¬w$ derive $¬w$ [with Disjunctive syllogism].
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
add a comment |
$begingroup$
Hint
With $lnot t$ and $z to t$ derive $lnot z$ [with Modus tollens].
With $lnot z$ and $w ∨ ¬z → r$ derive $r$ [with Addition and Modus ponens].
With $r$ and $lnot z$ and $¬z ∧ r → ¬s$ derive $lnot s$ [with Conjunction and Modus ponens].
With $lnot s$ and $s ∨ ¬w$ derive $¬w$ [with Disjunctive syllogism].
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
$endgroup$
Hint
With $lnot t$ and $z to t$ derive $lnot z$ [with Modus tollens].
With $lnot z$ and $w ∨ ¬z → r$ derive $r$ [with Addition and Modus ponens].
With $r$ and $lnot z$ and $¬z ∧ r → ¬s$ derive $lnot s$ [with Conjunction and Modus ponens].
With $lnot s$ and $s ∨ ¬w$ derive $¬w$ [with Disjunctive syllogism].
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
edited Jan 25 at 13:15
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
answered Jan 25 at 12:25
Mauro ALLEGRANZAMauro ALLEGRANZA
67.2k449115
67.2k449115
add a comment |
add a comment |
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