.

1. Prova por indu¸˜o:
ca ∀x(x ∈ N )(div7(23x − 1)).
a) Base da indu¸˜o ( div7(23×0 − 1)):
ca
1 ∀x(div7(x) ↔ ∃y(y ∈ N )(7y = x)) Defini¸˜o div7(x)
ca
2 23×0 − 1 = 0 Teorema
3 7×0=0 Teorema
4 7 × 0 = 23×0 − 1 2,3, = s
5 0∈N Teorema
6 ∃y(y ∈ N )(7y = 23×0 − 1) 4,5 I ∃
3×0
7 div7(2 − 1) 6,1 MP
b) Passo indutivo ( div7(23k − 1) → div7(23(k+1) − 1)):
1 ∀x(div7(x) ↔ ∃y(y ∈ N )(7y = x) Defini¸˜o div7(x)
ca
3k
2 div7(2 − 1) Hip´tese
o
3 ∃y(y ∈ N )(7y = 23k − 1) 2, 3, MP
3k
4 7a = 2 −1 4, E∃
5 7a + 1 = 23k 5, x + 1 = x + 1
3k
6 (7a + 1) × 8 = 2 ×8 6, 8x = 8x
7 (7a + 1) × 8 = 23k × 23 7, 23 = 8
8 (7a + 1) × 8 = 23k+3 8, xy × xz = xy+z
9 (7a + 1) × 8 = 23×(k+1) 9, Distributiva ×/+
3×(k+1)
10 8(7a) + 8 = 2 10, Distributiva ×/+
11 8(7a) + 7 = 23×(k+1) − 1 11, x − 1 = x − 1
3×(k+1)
12 7(8a) + 7 = 2 −1 12, Comutativa ×
13 7(8a + 1) = 23×(k+1) − 1 13, Distributiva ×/+
3×(k+1)
14 ∃y(7y = 2 − 1) 14, I∃
15 div7(23×(k+1) − 1) 15, 1, MP
16 div7(23k − 1) → div7(23×(k+1) − 1) 2, 16, I →
1