Examples of Logical Problems (Math Problem Sample)
set of logic problems. construct proofs and answer multiple choice questions.
source..I.
 1 ∀x(Ax → Bx)
 2 ∀x (Ax→Cx)
 3 Ad→Bd                                  1, ∀Elim
 4 Ad→Cd                                  2, ∀Elim
 5 Ad                                         Assumption for CP
 6 Bd                                           3,5 →Elim
 7 Cd                                      3,4 → Elim
 8 Bd & Cd                                  6,7 & Intro
 9 Ad → (Bd &Cd)                    5,-8  →Intro
 10 ∀x[Ax→(Bx&Cx)]                  9, ∀Intro
 1 ∀x(¬Cx →¬Ax)
 2 ∃xCx → (∃x¬(¬Bx V ¬ Dx)
 3 ¬(¬∃xAx V ∃xBx)                             Assumption for IP
 4 ¬Cd → ¬Ad                                 1, ∀ Elim
 5 Cd                                               Assumption for IP
 6 ∃xCx                                             5, ∃intro
 7 ∃x¬(¬Bx V ¬Dx)                            2,6 → elim
 8 ¬(¬Ba V ¬Da)                             Assumption for 7, E-elim
 9 ¬Ba                                           Assumption for IP
 10 ¬Ba V ¬Da                              9, V intro
 ⊥                                                         8, 10 ⊥ intro
 ¬¬Ba                                            9, 10-11 ¬ intro
 Ba                                             12, ¬ elim
 ∃xBx                                          13, ∃-intro
 ∃xBx                                          8, 9-11, ∃-elim
 ¬∃xAx V ∃xBx                              12, V intro
 ⊥                                                        3, 13 ⊥ intro
 ¬Cd                                                  5, 6-14 ¬intro
 ¬Ad                                               4, 16 → ∃ elim
 ∀x¬Ax                                           16, ∀ intro
 ∀x¬Ax                                           8, 9-20, ∃ elim
 Ab                                                 Assumption for IP
 ¬Ab                                           21, ∀ elim
 ⊥                                                       22, 23 ⊥ intro
 ¬∃x Ax                                            24, ⊥ elim
 ¬∃x Ax V ExBx                        25 V intro
 ⊥                                                  3, 26 ⊥ intro
 ¬¬(¬∃x Ax V ∃x Bx)              3, 4-27 ¬intro
 ¬∃x Ax V ∃x Bx                       28, ¬ Elim
 1 ¬∃x Ax V ∀x (¬Bx V Ex)
 2 ∃xBx → ∀ x¬Ex
 3 ¬∃x Ax                                                             Assumption for V elim
 4 Ab & Db                                                            Assumption for CP
 5 Ab                                                                 4, & Elim
 6 ∃x Ax                                                           5, ∃ intro
 7 ⊥                                                                         3, 6 ⊥ intro
 8 ¬Bb                                                               7, ⊥ elim
 9 Ab & Db  → ¬Bd                                  4-8  → intro
 10 ∀ x[(Ax&Dx) → ¬Bx]                               9, ∀ intro
 ∀ x(¬Bx V Ex)                                                 Assumption for V elim
 ¬Bd V Ed                                                     11, ∀ elim
 ¬Bd                                                                assumption for V elim
 Ab & Db  → ¬Bd                                    4-13, → intro
 ∀ x[(Ax & Dx) → ¬Bx]                               14, ∀ intro
 Ed                                                           assumption for V elim
 Dd                                                          4, & elim
 ∃x Dx                                                     17, ∃ intro
 ∀ x¬Ex                                                     2, 18  → elim
 ¬Ed                                                       19, ∀ elim
 ⊥                                                                     16, 20 ⊥ intro
 ¬Bd                                                           21, ⊥ elim
 Ab & Db  → ¬Bd                                       4-22  →intro
 ∀ x[(Ax & Dx) → ¬Bx]                          23, ∀ intro
 ∀ x[(Ax & Dx) → ¬Bx]                            12, 13-15, 16-24, V elim
 ∀ x[(Ax & Dx) → ¬Bx]                              1, 3-10, 11-25 V elim
 1 ∀ xAx ↔∃x (Bx & Cx)
 2 ∀ x (Cx   →Bx)
 3 ∀ xAx                                                 assumption for → intro
 4 ∃x (Bx & Cx)                                     ↔ elim
 5 Bd & Cd                                          4, ∃ elim
 6 Cd                                                   5, & elim
 7 ∃x Cx                                              6, ∃ intro
 8 ∃x Cx                                              4, 5-7 ∃ elim
 9 ∀ xAx →∃x Cx                                        3-7 → intro
 10 ∃x Cx                                               assumption for IP
 Cb                                                       9, ∃ elim
 Cb →Bb                                              2, ∀ elim
 Bb                                                  10, 11 → elim
 Bb & Cb                                       10, 12 & intro
 ∃x (Cx & Bx)                                       13, ∃ intro
 ∃x (Cx & Bx)                                       10, 11-15 ∃ elim
 ∀xAx                                                 1,16↔ elim
 ∃x Cx → ∀ xAx                                             9, 17 → intro
 ∀xAx ↔ ∃x Cx                                     8, 18, ↔ intro
 1 ∃x (¬Ax →Bx)
 2 ∃x Ax → Vx(Cx→Bx)
 3 ∃x Cx
 4 ¬ExBx                                                        assumption  for Ip
 5 Cd                                                                 3, ∃ elim
 6 ¬Ae→Be                                                    1, assumption for ∃ elim
 7 ¬Ae                                                          assumption for IP
 8 Be                                                        6, 7 → elim
 9 ∃x Bx                                                8, ∃ intro
 10 ∃x Bx                                                1, 6-9, ∃ Elim
 ⊥                                                                  4, 10 ⊥ intro
 ¬¬Ae                                                                7, 8-11 ¬ intro
 ∃x Ax                                                                                  12, ∃ intro
 Vx(Cx→Bx)                                                            2, 13 → elim
 Cd→Bd                                                                            14, ∀ elim
 Bd                                                                     5, 15 → elim
 ∃x Bx                                                         16, ∃ intro
 ⊥                                                    4, 17 ⊥ intro
 ¬¬∃x Bx                                       4, 5-18 ¬ intro
 ∃x Bx                                                19, ¬ elim
II.
W is a world in which all senators are republican and in which all democrats are not governors
D={a,b,c}
S={a,b}
D{c}
G{a}
III.
D={a,b}
A={a,b}
B{0-empty}
IV.
 1 D=(c,d,b}
F={0-empty]
 2 D={c,d,b,e}
K={,b,e>}
 3 D={c,d,e}
F={c,d,e}
G={c,}
 4 D={c,d,e}
F={0-emopty}
G={c,d,e}
 5 D={c,d,e}
F={c,d}
Q={e}
V. Not clear what above statement is and the requirement for an answer to this question
VI.
Statement 3 is not a logical truth because Txy can be given an interpretation on which the sentence is false: Txy – x likes y.
In a world in which Bob likes Jane and Jane likes Tom but in which Bob does not like Tom the sentence wo...
Other Topics:
- Solving Mathematical Equation Involving IntegrationDescription: To evaluate ʃ x sin 3x dx, we use the method of integration by parts given by the formula; ʃ u dv = uv-ʃ v du Let u = x, dv = sin 3x; then, v= -cos 3x and du = dx. Now substituting in the formula we get; ʃ x sin 3x dx = x-1/3cos 3x-ʃ-1/3cos 3x dx = -1/3cos 3x+1/9sin 3x+C, where C is a constant. ...2 pages/≈550 words| No Sources | Other | Mathematics & Economics | Math Problem |
 - Mathematics exam setting. Mathematics & Economics Math ProblemDescription: 5x x 52y = 25 and 32x x 3y = QUOTE (3 Marks) Solve for x in (log x)2 = QUOTE (3 marks) Solve for x in the equation. (3 marks) 9(22x+ 2) – 41(2x) + 8 = 0 Solve for x in the equation. EMBED Equation.3 (3 marks) Solve for x in the given equation. (3 marks) 64x – 121 = 7...1 page/≈550 words| 1 Source | Other | Mathematics & Economics | Math Problem |
 - Mathematics Guideline on the Warehouse Economics Math ProblemDescription: 572.31 square feet = 82944 cubic inch Give the need to store 200000 pallets within the warehouse, and there is a need for theoretical square feet measurements as given below: 572.31×200000=114462000...1 page/≈275 words| 2 Sources | Other | Mathematics & Economics | Math Problem |