The document covers assignments on the theory of inference for statement calculus, detailing various premises and conclusions to analyze logical validity. It includes showing that conclusions follow from given premises, determining the validity of conclusions based on multiple arguments, and deriving statements using rules such as the conditional proof rule. Additionally, it addresses inconsistencies among premises and employs indirect methods for certain proofs.

1. DMS Sunita M Dol
Walchand Institute of Technology, Solapur Page 1
Assignment No. 6
Topic covered:
• Theory of inference for statement calculus.
1. Show that the conclusion C follows from the premises H1, H2, … in the following cases
a. H1 : P → Q C : P → (P ∧ Q)
b. H1 : ¬P ∨ Q H2 : ¬(Q ∧ ¬P) H3: ¬R C : Q
c. H1 : ¬P H2 : P ∨ Q C : Q
d. H1 : ¬Q H2 : P → Q C : ¬P
e. H1 : P → Q H2 : Q → R C : P → R
f. H1 : R H2 : P ∨ ¬P C : R
2. Determine whether the conclusion C is valid in the following when H1, H2,... are
premises
a. H1 : P → Q H2: ¬Q C : P
b. H1 : P ∨ Q H2 : P → R H3: Q → R C : R
c. H1 : P → (Q → R) H2 : P ∧ Q C : R
d. H1 : P → (Q → R) H2 : R C : P
e. H1 : ¬P H2 : P ∨ Q C : P ∧ Q
3. Show the validity of the following arguments for which the premises are given on the left
and conclusion on right.
a. ¬ (P ∧ ¬Q), ¬Q ∨ R, ¬R ¬P
b. (A → B) ∧ (A → C), (B ∧ C), D ∨ A D
c. ¬J → (M ∨ N), (H ∨ G) → ¬J, H ∨ G M ∨ N
d. P → Q, (¬Q ∨ R) ∧ ¬R, ¬ (¬P ∧ S) ¬S
e. (P ∧ Q) → R, ¬R ∨ S, ¬S ¬P ∨ ¬Q
f. P → Q, Q → ¬R, R, P ∨ (J ∧ S) J S
g. B ∧ C, (B ↔ C) → (H ∨ G) G ∨ H
h. (P → Q) → R, P ∧ S, Q ∧ T R
4. Derive the following using CP rule
a. ¬P ∨ Q, ¬Q ∨ R, R → S ⇒P → S
b. P, P → (Q → (R ∧ S)) ⇒Q → S
c. P → Q ⇒P → (P ∧ Q)
d. (P ∨ Q) → R ⇒(P ∧ Q) → R
e. P → (Q → R), Q → (R → S) ⇒P → (Q → S)

2. DMS Sunita M Dol
Walchand Institute of Technology, Solapur Page 2
5. Show that following premises are inconsistent
a. P → Q, P → R, Q → ¬R, P
b. A → (B → C), D → (B ∧ ¬C), A ∧ D
6. Show the following (Use indirect method if needed)
a. (R → ¬Q), R ∨ S, S → ¬Q, P → Q ⇒¬P
b. S → ¬Q, S ∨ R, ¬R, ¬R ↔ Q ⇒¬P
c. ¬ (P → Q) → ¬ (R ∨ S), ((Q → P) ∨ ¬R), R ⇒P ↔ Q