Do the following proofs using only inference rules and replacement rules.

 

1. [ solution ]

1. p->~p /.:~p

 

2. [ solution ]

 

1. ~avy
2. (zvw)->(a&~y) /.:~z

 

3. [ solution ]

1. a->b
2. c->~b
3. d->(a&c)  /.:~d

 

4. [ solution ]

1. ~k->~m
2. ~m->l
3. k->~l /.:k<->m

 

 

5. [ solution ]

 

1. ~(avb)
2. a<->c
3. {(dva)<->e}->(cvb)
4. e->(avc) /.:d