Exercise 4.1.4 Prove the remaining identities in Proposition 4.1.6. (Hint: one can save some work by using some identities to prove others. For instance, once you know that \(xy = yx\), you get for free that \(x1 = 1x\), and once you also prove \(x(y + z) = xy + xz\), you automatically get \((y + z)x = yx + zx\) for free.)
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\(\bullet\) Show that \(x + y = y + x\).
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\(\Vdash\) \(x + y\)
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\(=\) { \(x = a — b\), \(y = c — d\) }
- \((a — b) + (c — d)\)
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\(=\) { Definition 4.1.2 }
- \((a + c) — (b + d)\)
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\(=\) { Commutativity of the addition on \(\mathbb N\) }
- \((c + a) — (d + b)\)
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\(=\) { Definition 4.1.2 }
- \((c — d) + (a — b)\)
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\(=\) { \(x = a — b\), \(y = c — d\) }
- \(y + x\)
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\(\square\)
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\(\bullet\) Show that \((x + y) + z = x + (y + z)\).
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\(\Vdash\) \((x + y) + z\)
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\(=\) { \(x = a — b\), \(y = c — d\), \(z = e — f\) }
- \(((a — b) + (c — d)) + (e — f)\)
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\(=\) { Definition 4.1.2 }
- \(((a + c) — (b + d)) + (e — f)\)
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\(=\) { Definition 4.1.2 }
- \(((a + c) + e) — ((b + d) + f)\)
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\(=\) { Associativity of the addition on \(\mathbb N\) }
- \((a + (c + e)) — (b + (d + f))\)
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\(=\) { Definition 4.1.2 }
- \((a — b) + ((c + e) — (d + f))\)
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\(=\) { Definition 4.1.2 }
- \((a — b) + ((c — d) + (e — f))\)
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\(=\) { \(x = a — b\), \(y = c — d\), \(z = e — f\) }
- \(x + (y + z)\)
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\(\square\)
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\(\bullet\) Show that \(x + 0 = 0 + x = x\).
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\(-\) Let \(c\in \mathbb N\) and \(0 := c — c\)
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\(\Vdash\) \(x + 0\)
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\(=\) { \(x = a — b\), \(0 = c — c\) }
- \((a — b) + (c — c)\)
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\(=\) { Definition 4.1.2 }
- \((a + c) — (b + c)\)
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\(=\) { Definition 4.1.1, Commutativity and associativity of the addition on \(\mathbb N\), \((a + c) + b = a + (b + c)\) }
- \(a — b\)
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\(=\) { \(x = a — b\) }
- \(x\)
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\(=\) { \(x = a — b\) }
- \(a — b\)
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\(=\) { Definition 4.1.1, Commutativity and associativity of the addition on \(\mathbb N\), \(a + (c + b) = (c + a) + b\) }
- \((c + a) — (c + b)\)
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\(=\) { Definition 4.1.2 }
- \((c — c) + (a — b)\)
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\(=\) { \(x = a — b\), \(0 = c — c\) }
- \(0 + x\)
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\(\square\)
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\(\bullet\) (Commutative law of \(\mathbb Z\)) Show that \(xy = yx\).
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\(\Vdash\) \(xy\)
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\(=\) { \(x = a — b\), \(y = c — d\) }
- \((a — b)(c — d)\)
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\(=\) { Definition 4.1.2 }
- \((ac + bd) — (ad + bc)\)
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\(=\) { Commutativity of the addition and the multiplication on \(\mathbb N\) }
- \((ca + db) — (cb + da)\)
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\(=\) { Definition 4.1.2 }
- \((c — d)(a — b)\)
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\(=\) { \(x = a — b\), \(y = c — d\) }
- \(yx\)
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\(\square\)
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\(\bullet\) Show that \(x1_Z = 1_Zx = x\).
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\(-\) Let \(1_Z = 1_N — 0_N\), \(x = a — b\)
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\(\Vdash\) \(x1_Z\)
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\(=\) { \(1_Z = 1_N — 0_N\), \(x = a — b\) }
- \((a — b)(1_N — 0_N)\)
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\(=\) { Definition 4.1.2 }
- \((a1_N + b0_N) — (a0_N + b1_N)\)
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\(=\) { Multiplication on \(\mathbb N\) }
- \(a — b\)
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\(=\) { \(x = a — b\) }
- \(x\)
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\(=\) { \(x = a — b\) }
- \(a — b\)
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\(=\) { Multiplication on \(\mathbb N\) }
- \((1_Na + 0_Nb) — (1_Nb + 0_Na)\)
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\(=\) { Definition 4.1.2 }
- \((1_N — 0_N)(a — b)\)
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\(=\) { \(1_Z = 1_N — 0_N\), \(x = a — b\) }
- \(1_Zx\)
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\(\square\)
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\(\bullet\) (Left distributive law of \(\mathbb Z\)) Show that \(x(y + z) = xy + xz\).
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\(-\) Let \(x = a — b\), \(y = c — d\), \(z = e — f\)
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\(\Vdash\) \(x(y + z)\)
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\(=\) { \(x = a — b\), \(y = c — d\), \(z = e — f\) }
- \((a — b)((c — d) + (e — f))\)
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\(=\) { Definition 4.1.2 }
- \((a — b)((c + e) — (d + f))\)
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\(=\) { Definition 4.1.2 }
- \((a(c + e) + b(d + f)) — (a(d + f) + b(c + e))\)
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\(=\) { Commutative and Distributive law of \(\mathbb N\) }
- \(((ac + bd) + (ae + bf)) — ((ad + bc) + (af + be))\)
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\(=\) { Definition 4.1.2 }
- \(((ac + bd) — (ad + bc)) + ((ae + bf) — (af + be))\)
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\(=\) { Definition 4.1.2 }
- \((a — b)(c — d) + (a — b)(e — f)\)
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\(=\) { \(x = a — b\), \(y = c — d\), \(z = e — f\) }
- \(xy + xz\)
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\(\square\)
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\(\bullet\) \((y + z)x = yx + zx\).
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\(\Vdash\) \((y + z)x\)
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\(=\) { Commutative law of \(\mathbb Z\) }
- \(x(y + z)\)
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\(=\) { Left distributive law of \(\mathbb Z\) }
- \(xy + xz\)
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\(=\) { Commutative law of \(\mathbb Z\) }
- \(yx + zx\)
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\(\square\)