レポート課題第 4 回【5/17 10:20a.m. 締切】 次の各設問に答えよ．
(1)
(2)
(3)
(4)
解答.
f (P ) = f (Q) = 0 のとき，f (P ∨ Q), f (P → Q) の値を求めよ．
f (P ) = 0, f (Q) = 1 のとき，f (P ∧ Q), f (P → Q) の値を求めよ．
f (P ) = 1, f (Q) = 0 のとき，f (P ∧ Q), f (P ∨ Q) の値を求めよ．
f (P ) = f (Q) = 1 のとき，f (P ∧ Q), f (P ∨ Q), f (P → Q) の値を求めよ．
(1) f (P ) = f (Q) = 0 のとき，
f (P ∨ Q) = max{f (P ), f (Q)} = max{0, 0} = 0
f (P → Q) = max{1 − f (P ), f (Q)} = max{1 − 0, 0} = max{1, 0} = 1
(2) f (P ) = 0, f (Q) = 1 のとき，
f (P ∧ Q) = min{f (P ), f (Q)} = min{0, 1} = 0
f (P → Q) = max{1 − f (P ), f (Q)} = max{1 − 0, 1} = max{1, 1} = 1
(3) f (P ) = 1, f (Q) = 0 のとき，
f (P ∧ Q) = min{f (P ), f (Q)} = min{1, 0} = 0
f (P ∨ Q) = max{f (P ), f (Q)} = max{1, 0} = 1
(4) f (P ) = f (Q) = 1 のとき，
f (P ∧ Q) = min{f (P ), f (Q)} = min{1, 1} = 1
f (P ∨ Q) = max{f (P ), f (Q)} = max{1, 1} = 1
f (P → Q) = max{1 − f (P ), f (Q)} = max{1 − 1, 1} = max{0, 1} = 1
6

1G 定理 3.1(vii) 第 1 式 X U Y = Y U X ) U Z = X U (Y U Z

(P ∧ Q ) → ¬P ∨ ¬Q

x ∃y P (x, y).

u|u ̸∈ (X ∪ Y u ̸∈ (X ∪ Y ) ¬(u ∈ (X ∪ Y ¬(u ∈ X ∨ u