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Solve the following exercises using Gentzen's natural deduction method for predicate logic, applying the derived inference rules presented in proof theory:

LEVEL 1

[Exercise 19] [Exercise 20]

-1.∀x(Px→ Qx)

⊦∀x(Px→ Sx ∨ Rx) -1.∀xPx → ∀xQx ⊦¬∀xPx
-2.∀x(¬Sx→ ¬Qx)   -2. ¬Qa  
3. Pa      
| 4. Pa→ Qa E∀1    
| 5. Qa E→3,4    
| 6. ¬Sa → ¬Qa E∀2    
| 7. Sa MT 5,6    
8. Sa ∨ Ra I∨ 7    
9.Pa → Sa ∨ Ra I→3-8    
10. ∀x(Px→ Sx ∨ Rx) I∀ 10    

LEVEL 2

[Exercise 21] [Exercise 22]
-1.∀x(Tx → Mx) ⊦∀x(Tx→ Mx) -1. ∀xMx ⊦¬∀x∃y¬(Mx→¬Lxy)
-2. ∀x¬(Mx ∧ Rx)   -2. ∀x¬Lxx  
3. ∀x(Tx →( Px → Rx))   -3. ¬∃x∃y(Lxy ∧¬Lxx)  
    4,∀x∀y¬(Lxy ∧¬Lxx)  
    5. ¬(Lab ∧¬Laa)  
    6. Lab → Laa  
    7. ¬Laa  
    8. ¬Lab  
    9.Ma →¬Lab  
    10. ∀y(Ma →¬Lay)  
    11.∃x∀y(Mx →¬Lxy)  
    12. ∃x¬∃y¬(Mx →¬Lxy)  
    13. ∀x∃y¬(Mx→¬Lxy)  

LEVEL 3

[Exercise 23] [Exercise 24]
-1.∀x∀y∀z(¬(Txy→Txz) →¬Qyz) ⊦¬∃xRxa

-1.∀x(¬Fa ∨ Qx)

⊦∀x¬(Qx∧ Rx)→ Fa ∨ ¬Tbb)

-2.∀x∀y∀z(Rya→Qzx)   -2. ∀x(Qx ∧Txb → Rx)  
-3. ∃x∃y∃z(Txz ∧ Txy)      
4. Tbc ∧Tba      
| 5. Rba → Qca E∀2(y:b, x:a, z:c)    
| 6. ¬(Tbc → Tba) → ¬Qca E∀1(x:b, y:c, z:a)    
| 7. Qca → (Tbc →Tba) CTPS 6    
| 8. Rba →(Tbc →Tba) Tr → 5, 8    
| 9. Rba →¬(Tbc ∧¬Tba) Def (intersec) →/∧ 8    
| 10. ¬Rba MT 4,9    
| 11. ∀x¬Rba I∀11    
12. ¬∃xRxa Def ∀/∃ 11    
13. ¬∃xRxa E∃4-12    

 

Solutions

 

 

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