If p is true, q, r, and s are false, find the truth value of the proposition.
¬(¬𝑝 ∨ 𝑞) ∨ ¬((𝑝 ∧ ¬𝑟) → ¬𝑠)
let True is 1; False is 0; then\text {let True is 1; False is 0; then}let True is 1; False is 0; then
¬(¬𝑝∨𝑞)∨¬((𝑝∧¬𝑟)→¬𝑠)=¬(¬𝑝 ∨ 𝑞) ∨ ¬((𝑝 ∧ ¬𝑟) → ¬𝑠) =¬(¬p∨q)∨¬((p∧¬r)→¬s)=
¬(¬1∨0)∨¬((1∧¬0)→¬0)=¬(¬1 ∨ 0) ∨ ¬((1 ∧ ¬0) → ¬0)=¬(¬1∨0)∨¬((1∧¬0)→¬0)=
¬(0∨0)∨¬((1∧1)→1)=¬(0 ∨ 0) ∨ ¬((1 ∧ 1) → 1)=¬(0∨0)∨¬((1∧1)→1)=
¬(0)∨¬(1→1)=1∨¬(0)=1∨1=1;¬(0 ) ∨ ¬(1 → 1)=1∨ ¬(0)=1∨ 1= 1;¬(0)∨¬(1→1)=1∨¬(0)=1∨1=1;
¬(¬𝑝∨𝑞)∨¬((𝑝∧¬𝑟)→¬𝑠)=1 is True¬(¬𝑝 ∨ 𝑞) ∨ ¬((𝑝 ∧ ¬𝑟) → ¬𝑠)= 1 \text{ is True}¬(¬p∨q)∨¬((p∧¬r)→¬s)=1 is True
Answer: ¬(¬𝑝∨𝑞)∨¬((𝑝∧¬𝑟)→¬𝑠) is True¬(¬𝑝 ∨ 𝑞) ∨ ¬((𝑝 ∧ ¬𝑟) → ¬𝑠) \text{ is True}¬(¬p∨q)∨¬((p∧¬r)→¬s) is True
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