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Question 1104291: Translate and solve the proof:
If God believes on Monday that I'll tell a lie on Tuesday, then either I have the power to make one of God's past beliefs false, or I cannot refrain from lying on Tuesday. I do not have the power to make one of God's past beliefs false if either God is infallible or the past is unalterable. The past is unalterable. It follows that if God believes on Monday that I'll tell a lie on Tuesday, then I cannot refrain from lying on Tuesday. (B,F,R,I,P)
Click here to see answer by Alan3354(69443)

Question 1104306: P∙ Q
R⊃ ~Q
/ ~R∨ ~P

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Question 1104281: Use Mathematical Induction to show that the following statement is true for all natural numbers n:

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Question 1104538: . I had pancakes for breakfast today. I had pancakes for breakfast today and God infallibly foreknows the future, only if God believed a million years ago that I would have pancakes for breakfast today. If God believed a million years ago that I would have pancakes for breakfast today, then, if I could have chosen to have something else, either I could have brought it about that God did not believe a million years ago that I would have pancakes for breakfast today or I could have brought it about that God had a false belief. However, if God believed a million years ago that I would have pancakes for breakfast today, then, if I could have brought it about that God did not believe a million years ago that I would have pancakes for breakfast today, I could have changed the past. But I could not have changed the past. Also, it�s false that, if God infallibly foreknows the future, I could have brought it about that God had a false belief. Therefore, if God infallibly knows the future, I could not have chosen to have something else. (P: I had pancakes today; I: God infallibly foreknows the future; B: God believed a million years ago that I would have pancakes today; C: I could have chosen to have something else; Y: I could have brought it about that God did not believe a million years ago that I would have pancakes today; F: I could have brought it about that God had a false belief; H: I could have changed the past.

I have spent hours on this. You have to translate it then solve it by assuming the conditional proof or buy assuming RAA. I can't figure out the proper translation for it. Any help is very much appreciated I have a final exam in 1 hour.
Click here to see answer by Alan3354(69443)

Question 1104863: 1. P V Q
2. Q -> (R & S)
3. (R V P) -> T / T
Click here to see answer by Edwin McCravy(20054)

Question 1104861: 1. A -> (B -> C)
2. A -> B
3. ~ C ->(A V D) / C V D
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Question 1105010: Show:
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Question 1104862: 1. ~ A V (B & E)
2. ~ A -> ~ C
3. C V (B & D) / B
Click here to see answer by Edwin McCravy(20054)

Question 1104588: (AvB) v C, ~C, ~A, B�>D |- D
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Question 1104540: 1. N ⊃ (F � A)
2. B ⊃ (R � F) / (N ∨ B) ⊃ (A ∨ R)
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Question 1107829: For any two real numbers x and y, abs*(x-y)=abs*(y-x)
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Question 1109316: find the proof of the following:
P & Q, P -> R |- R & Q
(P v Q) -> R |- ~R-> ~Q
Click here to see answer by Edwin McCravy(20054)

Question 1108255: How do I complete the following proofs?
Premises:
1. L ⊃ M
2. L
3. K v Q
4. ~K Prove: M ∙ Q
Premises:
1. (A v ~B) v C
2. ~A Prove: B ⊃ C
Premises:
1. M v N
2. O ⊃ ~N Prove: O ⊃ M

Click here to see answer by Edwin McCravy(20054)

Question 1111798: If a,b,c, is even integer then the phythagorean theorem is even integer.
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Question 1111796: If x,y,z is odd integer then the result of this is odd integer by using quadratic formula..
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Question 1111800: For all a,b,c if a and b is odd and c is even then the sum and the product is odd integer...
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Question 1111799: If x,y,z is rational the the product is rational...
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Question 1111799: If x,y,z is rational the the product is rational...
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Question 1111797: If x,y,z is rational then the sum is rational..
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Question 1111797: If x,y,z is rational then the sum is rational..
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Question 1111996: Prove that a,b,c is even integer using pythagorean theorem
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Question 1111997: Prove that the sum of rational numbers is rational.
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Question 1112733: 1) A v B
2) B > (A v D)
3) ~D
Conclusion: A
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Question 1112710: Indirect proof
9.
1) R
2) (~ C v ~ D) v S
3) ~ (C ⋅ D) ⊃ ~R / ∴ S
10.
1) (A ⋅ B)⋅ ~ (S v T)
2) ~E
3) (S v T) v ~ (~E ⋅ ~F)
4) (~E v F)⊃(A ⋅ B) / ∴ E v F
12.
1) A v B
2) B ⊃ (A v D)
3) ~ D / ∴ A
Conditional proof
11.
1) A ⊃ B / ∴ A ⊃ [ C ⊃ ~ (B ⊃ ~A) ]
Help me please!
Click here to see answer by solver91311(24713)

Question 1112838: Simplify the lefthandside so that LHS=RHS
2/sin(b) = sin(b)/cos(b)-1 + sin(b)/cos(b)+1
This is a trig identity and I have no clue how to reach my goal on the left-hand side.

Click here to see answer by greenestamps(13198)

Question 1112729: 1) R
2) (~C v ~D) v S
3) ~(C & D) > ~R
Conclusion: S
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Question 1112731: 1) (A&B) > ~(S v T)
2) ~E
3) (S v T) v ~(~E & ~F)
4) (~E v F) > (A&B)
Conclusion: E v F
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Question 1113462: -2 and +2 are equal, if yes how ?
we have tried this-
(-2)x(-2)= 2x2
=> (-2)^2=2^2
=> (-2)=2
Is it correct ?
Click here to see answer by math_helper(2461)

Question 1113462: -2 and +2 are equal, if yes how ?
we have tried this-
(-2)x(-2)= 2x2
=> (-2)^2=2^2
=> (-2)=2
Is it correct ?
Click here to see answer by Alan3354(69443)

Question 1113462: -2 and +2 are equal, if yes how ?
we have tried this-
(-2)x(-2)= 2x2
=> (-2)^2=2^2
=> (-2)=2
Is it correct ?
Click here to see answer by ikleyn(52775)

Question 924485: Prove algebraically that
(2n+1)^2-(2n+1) is an even number
for all positive integer values of n.
Please help :)
Click here to see answer by amalm06(224)

Question 924485: Prove algebraically that
(2n+1)^2-(2n+1) is an even number
for all positive integer values of n.
Please help :)
Click here to see answer by ikleyn(52775)

Question 924485: Prove algebraically that
(2n+1)^2-(2n+1) is an even number
for all positive integer values of n.
Please help :)
Click here to see answer by MathTherapy(10551)

Question 1113758: I need help on this problem. I don't understand how to do proofs and I need all the help I can get. In the directions, it says, " Construct proofs to show the following symbolic arguments are valid. The comma marks the break between premises. " Here is the problem, that I need help on.
8. C-> (T->L), ~L, ~E->C, L v ~E :.~T
This the work I put on so far: 1. C -> (T ->L)
2. ~L
3. ~E->C
4. L v ~E
Click here to see answer by Edwin McCravy(20054)

Question 1113762: I need help on constructing the proof to make the argument valid. The Commas are the breaks. Also, I don't know if I am doing this right
9.~~A, B-> ~A, A :. ~B
My work:

1.~~A
2. B->~A
3. A
4. :. ~B
ADD Step 1, 2
MT Steps 3, 4
Am I correct? I don't know if I am doing this right?
Click here to see answer by math_helper(2461)

Question 1116612: 1. (A∨~B)→(F∨(R∙G))
2. A
3. F→L
4. (R∙G)→T
5. (L∨T)→S ∴ S
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Question 1116635: Write out the conclusion that follows in a single step from the given premises (please read U as horseshoe):
1. ~M U S
2. ~M
3. (M v H) v ~S
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Question 1116996: Construct a regular proof to derive the conclusion of the following argument:
1. X >Y
2. (Y v ~X) > (Y > Z) / ~Z > ~X
I have attempted to use Addition to get X > (Y V ~X) from line 1 and no proof checkers that I have attempted will accept it.
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Question 1117158: 6,12,30,56,132,?
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Question 1117150: Solve use reductio ad absurdum
1. ~P→(R∙S)
2. ~Q→(R∙T)
3. ~(S∨T) ∴ P∙Q
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Question 1116845: 1.~G
2.~I
3. ~H>(IvG)//H
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Question 1117149: Solve using the methods of Reductio ad Absurdum
1. ~A∙~B ∴ A↔B
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Question 1116905:
Give reasons for the steps, given premises 1, 2 and 3: 1. X ⊃ (Y ⊃ Z) 2. X ⊃ (A ⊃ B) 3. X � (Y ∨ A) ∴ ∼B ⊃ X 4. X 5. A ⊃ B 6. Y ⊃ Z 7. (Y ∨ A) � X 8. Y ∨ A 9. ∼(∼Y) ∨ A 10. ∼Y ⊃ A 11. ∼Y ⊃ B 12. ∼B ⊃ ∼(∼Y) 13. ∼B ⊃ Y 14. X ∨ ∼Y 15. ∼Y ∨ X 16. Y ⊃ X 17. ∼B ⊃ X

Click here to see answer by Edwin McCravy(20054)

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