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Truth value of a compound formula (a)

Short answer

S→(P∧¬R) evaluates to false (T→F), while ((P→(R∨Q))∧S is true. The conjunction is false; truth value: FALSE.

If P = F, Q = T, S = T, R = F, then find truth value of: (S → (P ∧ ¬R)) ∧ ((P → (R ∨ Q)) ∧ S)

Given P = F, Q = T, S = T, R = F. Formula: (S → (P ∧ ¬R)) ∧ ((P → (R ∨ Q)) ∧ S)

Part 1: S → (P ∧ ¬R)

  • ¬R = ¬F = T
  • P ∧ ¬R = F ∧ T = F
  • S → (P ∧ ¬R) = T → F = F

Part 2: (P → (R ∨ Q)) ∧ S

  • R ∨ Q = F ∨ T = T
  • P → (R ∨ Q) = F → T = T
  • (P → (R ∨ Q)) ∧ S = T ∧ T = T

Whole formula

F ∧ T = F

Sub-formulaValue
¬RT
P ∧ ¬RF
S → (P ∧ ¬R)F
R ∨ QT
P → (R ∨ Q)T
(P → (R ∨ Q)) ∧ ST
Whole formulaF

Answer: the truth value is FALSE.

✓ Answer checked by Claude Opus 5.5 (high effort)

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