Posted: February 10th, 2022

7.9.2. Construct formal proofs for all the arguments below. Use equivalence rules, truth functional arguments, and the rules of instantiation and generalization. These may also be proven using the method of tableaux.

6. ∀x(Cx ⊃ ¬Sx), Sa ∧ Sb ∴ ¬(¬Ca ⊃ Cb)

7. ∃xCx ⊃ ∃x(Dx ∧ Ex), ∃x(Ex ∨ Fx) ⊃ ∀xCx ∴ ∀x(Cx ⊃ Gx)

8. ∀x(Fx ⊃ Gx), ∀x[(Fx ∧ Gx) ⊃ Hx] ∴ ∀x(Fx ⊃ Hx)

9. ∃xLx ⊃ ∀x(Mx ⊃ Nx), ∃xPx ⊃ ∀x ¬Nx ∴ ∀x[(Lx ∧ Px) ⊃ ¬Mx]

10. ∀x(Fx ≡ Gx), ∀x[(Fx ⊃ (Gx ⊃ Hx)], ∃xFx ∨ ∃xGx ∴ ∃xHx

11. ∃x(Cx ∨ Dx), ∃xCx ⊃ ∀x(Ex ⊃ Dx), ∃xEx ∴ ∃xDx

12. ∀x[(¬Dx ⊃ Rx) ∧ ¬(Dx ∧ Rx)], ∀x[Dx ⊃ (¬Lx ⊃ Cx)], ∀x(Cx ⊃ Rx) ∴ ∀x(Dx ⊃ Lx)

ASSIGNMENT 5:

7.9.3. Using the method of tableaux, give an assignment of values for the predicates of each argument that shows that each argument is invalid.

1. ∀x(Ax ⊃ Bx), ∀x(Ax ⊃ Cx) ∴ ∀x(Bx ⊃ Cx)

2. ∃x(Ax ∧ Bx), ∀x(Cx ⊃ Ax) ∴ ∃x(Cx ∧ Bx)

3. ∀x[(Cx ∨ Dx) ⊃ Ex], ∀x[(Ex ∧ Fx) ⊃ Gx] ∴ ∀x(Cx ⊃ Gx)

4. ∃xMx, ∃xNx ∴ ∃x(Mx ∧ Nx)

5. ∀x[Dx ∨ (Ex ∨ Fx)] ∴ ∀xDx ∨ (∀xEx ∨ ∀xFx)

6. ∃x(Cx ∧ ¬Dx), ∃x(Dx ∧ ¬Cx) ∴ ∀x(Cx ∨ Dx)

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