This assignment consists of 4 parts (i.e., Parts A– D). Total possible high score = 100 marks.

Part A: Symbolic Translation.

Translate the following ordinary language statements into predicate logic using the suggested notation.
(4 marks each. Total = 20 marks.)

  1. A thing is a cat only if it is an animal.
  2. Every logic student is sensible.
  3. Everything is an illusion.
  4. Nothing is right unless it is not wrong.
  5. If everything is mental, then nothing is physical.

(Ax: x is an animal; Cx: x is a cat; Lx: x is a logic student; Sx: x is sensible; Ix: x is an illusion; Rx: x is right; Wx: x is wrong; Mx: x is mental; Px: x is physical.)

Part B: Symbolic Translation.

Translate the following ordinary language argument into predicate logic using the suggested notation.
(2 marks each. Total = 20 marks)

Some criminal robbed the Russell mansion. Whoever robbed the Russell mansion either had an accomplice among the servants or had to break in. To break in, one would either have to smash the door or pick the lock. Only an expert locksmith could have picked the lock. Had anyone smashed the door, he would have been heard. Nobody was heard. If the criminal who robbed Russell mansion managed to fool the guard, he must have been a convincing actor. No one could rob the Russell mansion unless he fooled the guard. No criminal could be both an expert locksmith and a convincing actor. Therefore, some criminal had an accomplice among the servants.

(Cx: x is a criminal; Rx: x robbed the Russell mansion; Sx: x had an accomplice among the servants; Bx: x broke in; Dx: x smashed the door; Px: x picked the lock; Lx: x is an expert locksmith; Hx: x was heard; Fx: x fooled the guard; Ax: x is a convincing actor.)

Part C: Proof of validity.

The following 4 arguments are valid. Provide a proof of validity for each argument. You can use the 18 rules of inference, CP, IP, the rules for introducing and removing quantifiers and the Change of Quantifier rules.
(10 marks each. Total = 40 marks)

    1. (x) ~ Bx
    2. Jc ⊃ Bc ∴(∃ x) ~ Jx

 

    1. (x) [(Ax ⊃ Bx)
    2. (∃ x) Ax ∴(∃ x) (Ax • Bx)

 

    1. (∃x) Fx ⊃ (∃x) (Gx & Hx)
    2. (∃x) Hx ⊃ (x) Jx ∴(x) (Fx ⊃ Jx)

 

    1. ~ [(x) Ax ⊃ (∃ x) Bx]
    2. (∃x) ~Cx ⊃ (∃x) Bx
    3. (x) [( Ax • Cx ) ⊃ Da ∴Da

Part D: Proving Invalidity.

The following two arguments are invalid. Show them to be invalid using the finite universe method.
(10 marks each. Total = 20 marks)

    1. (x) (Mx ⊃ Nx)
    2. (∃ x) (Mx) ∴(x) Nx

 

    1. (∃ x) (Ax & Bx)
    2. (∃ x) (Cx • Bx) ∴ (∃ x) (Ax • Cx)

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