josephmary said:
i came acroos the below while studying propositional Logic, can anyone find the proofs
1) P ⊢ P
2) P → Q, Q→R ⊢ P → R
3) P → Q, Q→R, ¬R ⊢ ¬P
4) Q→R ⊢ (PvQ) → (PvR)
5) P →Q ⊢ (P&R) → (Q&R)
1)$P$..................Assumption
2)$\neg P$................Hypothesis for contradiction
3)$P\wedge\neg P$............(1),(2) and using addition Introdaction
4)$\neg\neg P$..................From (2) to (3) and using contradiction
5) $P$..................(4) negation elimination
(2) and (3) are easy to do ,you can use hypothetical syllogism for (2) or conditional proof and modus ponens
And hypothetical syllogism , contrapositive and modus ponens for (3) or contradiction,and modus ponens
I will do (4) :
1)$Q\Rightarrow R$..............Assumption
2)$P\vee Q$..................Hypothesis for conditional proof
3)$\neg(P\vee R)$................Hypothesis for contraction
4)$(\neg P\wedge\neg R)$............From (3) and using de Morgan
5)$\neg P$..................(4), Addition elimination (AE)
6)$\neg R$..................(4),AE
7)$\neg R\Rightarrow\neg Q$.............(1),Contrapositive
8)$\neg Q$..................(6),(7),Modus Ponens(MP)
9)$\neg P\Rightarrow Q$..............(2),material implication
10)$Q$.....................(5),(9) MP
11)$Q\wedge\neg Q$................(8),(10) Addition Introduction (AI)
12)$\neg\neg(P\vee R)$...............from (3) to (11) and using contradiction
13)$(P\vee R)$...................(12),negation elimination
14)$(P\vee Q)\Rightarrow(P\vee R)$............from (2) to (13) and using conditional proof
(5) is on the same style with (4) and even easier
