k
(d)
[p]1
[¬(p ∨ ¬p)]3 p ∨ ¬p
⊥
¬p 1
[¬p]2 [¬(p ∨ ¬p)]3 p ∨ ¬p
⊥
¬¬p 2
⊥
p ∨ ¬p 3
(f) ND (( p → q ) ∧ ( p → ¬ q )) → ¬ p :
[p]1
[(p → q) ∧ (p → ¬q)]2
p → q q
[p]1
[(p → q) ∧ (p → ¬q)]2
p → ¬q
¬q
⊥
¬p 1 ((p → q) ∧ (p → ¬q)) → ¬p 2
[A]1
⊥
¬A 1
Hence ¬ B ND¬ A (b) Suppose ¬ B ND¬ A
[¬B]1
⊥
B 1
Hence A NDB