CS计算机代考程序代写 137 Let a , b , and c be integer variables. Express as simply as possible without using quantifiers, assignments, or dependent compositions

137 Let a , b , and c be integer variables. Express as simply as possible without using quantifiers, assignments, or dependent compositions
(a) §
(b) §
(c) §
(d) §
(e) §
(f) §
= = = =
= = = =
= = = = = = =
= = = =
= =
= = =
= = = = =
b:= a–b. b:= a–b
b:= a–b. b:= a–b
b:= a–b. aʹ=a ∧ bʹ = a–b ∧ cʹ=c aʹ=a∧bʹ=a–(a–b)∧cʹ=c
aʹ=a ∧ bʹ=b ∧ cʹ=c
ok
a:= a+b. b:= a–b. a:= a–b
a:= a+b. b:= a–b. a:= a–b
a:= a+b. b:= a–b. aʹ = a–b ∧ bʹ=b ∧ cʹ=c a:= a+b. aʹ = a–(a–b) ∧ bʹ = a–b ∧ cʹ=c aʹ=b ∧ bʹ = (a+b)–b ∧ cʹ=c
aʹ=b ∧ bʹ=a ∧ cʹ=c
expand last assignment Substitution Law
expand last assignment Substitution Law Substitution Law subtract
(g)
aʹ = a+b+1. bʹ = a–b–1
c:= a–b–c. b:= a–b–c. a:= a–b–c. c:= a+b+c
c:= a–b–c. b:= a–b–c. a:= a–b–c. c:= a+b+c
c:= a–b–c. b:= a–b–c. a:= a–b–c. aʹ=a ∧ bʹ=b ∧ cʹ=a+b+c c:= a–b–c. b:= a–b–c. aʹ=a–b–c ∧ bʹ=b ∧ cʹ=(a–b–c)+b+c c:= a–b–c. b:= a–b–c. aʹ=a–b–c ∧ bʹ=b ∧ cʹ=a
c:= a–b–c. aʹ=a–(a–b–c)–c ∧ bʹ=a–b–c ∧ cʹ=a
c:= a–b–c. aʹ=b ∧ bʹ=a–b–c ∧ cʹ=a
aʹ=b ∧ bʹ=a–b–(a–b–c) ∧ cʹ=a
aʹ=b ∧ bʹ=c ∧ cʹ=a
a:= a+b. b:= a+b. c:= a+b
a:= a+b. b:= a+b. c:= a+b
a:= a+b. b:= a+b. aʹ=a ∧ bʹ=b ∧ cʹ = a+b a:= a+b. aʹ=a ∧ bʹ = a+b ∧ cʹ = a+a+b
aʹ = a+b ∧ bʹ = a+b+b ∧ cʹ = a+b+a+b+b aʹ=a+b ∧ bʹ = a + 2×b ∧ cʹ = 2×a + 3×b
expand last assignment Substitution Law arithmetic Substitution Law arithmetic Substitution Law arithmetic
expand last assignment substitution law substitution law arithmetic
a:= a+b. bʹ = a+b. c:= a+b
a:= a+b. bʹ = a+b. c:= a+b
a:= a+b. bʹ = a+b. aʹ=a ∧ bʹ=b ∧ cʹ = a+b
a:= a+b. ∃aʹʹ, bʹʹ, cʹʹ· bʹʹ = a+b ∧ aʹ=aʹʹ ∧ bʹ=bʹʹ ∧ cʹ = aʹʹ+bʹʹ
a:= a+b. bʹ = a+b ∧ cʹ = aʹ+bʹ bʹ = a+b+b ∧ cʹ = aʹ+bʹ
bʹ = a + 2×b ∧ cʹ = aʹ+bʹ
expand last assignment dependent composition
one-point for aʹʹ and bʹʹ , idempotence for cʹʹ substitution law arithmetic
a:= a+b+1. b:= a–b–1. a:= a–b–1
a:= a+b+1. b:= a–b–1. a:= a–b–1
a:= a+b+1. b:= a–b–1. aʹ = a–b–1 ∧ bʹ=b ∧ cʹ=c a:= a+b+1. aʹ = a–(a–b–1)–1 ∧ bʹ=a–b–1 ∧ cʹ=c a:= a+b+1. aʹ=b ∧ bʹ=a–b–1 ∧ cʹ=c
aʹ=b ∧ bʹ=a+b+1–b–1 ∧ cʹ=c
aʹ=b ∧ bʹ=a ∧ cʹ=c
expand last assignment substitution law once simplify substitution law simplify

§
=
aʹ = a+b+1. bʹ = a–b–1
∃aʹʹ, bʹʹ, cʹʹ· aʹʹ = a+b+1 ∧ bʹ = aʹʹ–bʹʹ–1 ∃bʹʹ· bʹ = a+b+1–bʹʹ–1
∃bʹʹ· bʹʹ = a+b–bʹ ∧ T
expand dependent composition one point for aʹʹ , identity for cʹʹ
simplify, rearrange, and identity one point for bʹʹ
expand last := substitution law substitution law
simplify
(h) §
= = = =
a:= a–b. b:= a–b. a:= a+b
a:= a–b. b:= a–b. a:= a+b
a:= a–b. b:= a–b. aʹ = a+b ∧ bʹ=b ∧ cʹ=c a:= a–b. aʹ = a+a–b ∧ bʹ=a–b ∧ cʹ=c
aʹ = a–b+a–b–b ∧ bʹ=a–b–b ∧ cʹ=c
aʹ = 2×a – 3×b ∧ bʹ = a – 2×b ∧ cʹ=c
=
= =T