CSCE 222 CSCE222 Homework 1 – Texas A&M

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CSCE 222 CSCE222 Homework 1 – Texas A&M

Homework 1

Due: 8 September 2017 (Friday) before 11:59 p.m.on eCampus (ecampus.tamu.edu).You must show your work in order to recieve credit.

Problem 1.(20 points)Use propositional equivalences to determine the correct equivalence for each expression:

1. (¬¬¬(p⊕p)∨ ¬¬((¬(p∧p)∧ ¬(p⊕ ¬¬p))→ ¬¬¬p))∧ ¬¬¬p≡

(a)T

(b)F

(c)p

(d)¬p

SOLUTION

(¬¬¬(p⊕p)∨ ¬¬((¬(p∧p)∧ ¬(p⊕ ¬¬p))→ ¬¬¬p))∧ ¬¬¬p

≡(¬¬¬(p)∨ ¬¬((¬(p)∧ ¬(p)→ ¬p))∧ ¬pby Idempotence & double negation≡ ¬¬¬p∨ ¬¬((¬p)→ ¬p)∧ ¬pby Idempotence≡ ¬¬¬p∨ ¬¬(p∨ ¬p)∧ ¬pby ((¬p→ ¬p)≡(p∨ ¬p))

≡ ¬p∨(p∨ ¬p)∧ ¬pby DeMorgans & double negation

≡ ¬p∨T∧ ¬pby Negation law≡ ¬(p∧F)∧ ¬pby inverse DeMorgan

≡ ¬(F)∧ ¬pby domination law

≡T∧ ¬pby¬F

≡ ¬(F∨p) by inverse DeMorgan

≡ ¬pby Identity law

(¬¬¬(p⊕p)∨ ¬¬((¬(p∧p)∧ ¬(p⊕ ¬¬p))→ ¬¬¬p))∧ ¬¬¬p≡ ¬p

2. (¬¬(¬¬p⊕ ¬(p∨p))↔ ¬(¬((p↔p)→ ¬p)→(¬(¬p∨p)↔ ¬(p⊕p))))↔ ¬(p→p)≡

(a)T

(b)F

(c)p

(d)¬p

SOLUTION

(¬¬(¬¬p⊕ ¬(p∨p))↔ ¬(¬((p↔p)→ ¬p)→(¬(¬p∨p)↔ ¬(p⊕p))))↔ ¬(p→p)

≡(¬¬(p⊕ ¬p)↔ ¬(¬((p↔p)→ ¬p)→(¬(¬p∨p)↔ ¬(p⊕p))))↔ ¬(p→p) by doublenegation & Idempotence≡ ¬¬T↔ ¬(¬((p↔p)→ ¬p)→(¬(¬p∨p))↔ ¬(p⊕p)↔ ¬(p→p) byp⊕ ¬p≡p

≡ ¬¬T↔ ¬(¬((p↔p)→ ¬p)→(p∧negp)↔ ¬(p⊕p))↔ ¬(p→p) by DeMorgans Law

≡ ¬¬T↔ ¬(¬((p↔p)→ ¬p)→F↔ ¬(p⊕p))↔ ¬(p→p) by Negation

≡T↔ ¬(¬((p↔p)→ ¬p)→F↔ ¬(F))↔ ¬(p→p) by double negation law $p⊕p≡F

≡T↔ ¬(¬(((p→p)∨(p→p))→ ¬p→F↔ ¬(F)))↔ ¬(p→p) byp↔p≡(p→p)∧(p→p)

≡T↔ ¬(¬(T→ ¬p)→F↔ ¬(F))↔ ¬(p→p) byp→p≡ ¬p∧p≡T&T∨T≡T

≡T↔ ¬(¬(¬p)→(F↔ ¬F)↔ ¬(p→p) byT→ ¬p≡F∧ ¬p≡ ¬p

≡T↔ ¬(p→(F↔ ¬F))↔ ¬(p→p) by double negation≡T↔ ¬(p→F)↔ ¬(p↔p) by (F→ ¬F)∧(¬F→F)≡F

≡T↔ ¬(¬p)↔ ¬(p→p) byp→F≡ ¬p∨F≡ ¬(p∧T)≡ ¬p

≡T↔p↔ ¬(p→p) by double negation≡p↔ ¬(p→p) by (T→p)∧(p→T)≡p≡p↔ ¬(¬p∨p) byp→p≡ ¬p∨p≡p↔Fby Negation law≡ ¬pbyp↔F≡(p∨F)≡p(¬¬(¬¬p⊕ ¬(p∨p))↔ ¬(¬((p↔p)→ ¬p)→(¬(¬p∨p)↔ ¬(p⊕p))))↔ ¬(p→p)≡p

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