回應余創豪君對梁、李對答之討論
張國棟
也許生不逢時，又或當年仍年少無知，我對梁燕城、李天命之間的爭論從不
感覺「埋身」，故亦沒有細讀那大堆文字。然而單就余君在「從研究方法論看護
教學」一文提及《哲客俠情》中的一個論證，並指出其中並沒有李所指的方法論
問題，我卻有點保留。
我認為就算沒有隱藏假設，梁的講法仍是不能成立的。假設天下間只有a=
「造一塊沒有人能舉起的石頭」，和b=「舉起一塊任何重量的石頭」兩種活動，
並假設它們足以定義何謂「全能」，那麼若X可以a來描述，X就是「全能」，
又或者若Y可以b來描述，X就是「全能」。
(1) 若X能做a，則X是全能者。
(2) 若Y能做b，則Y是全能者。然而問題的核心並非 (1) 和 (2) 是否成
立，而是X可否就是Y？
(3) 由於這可能世界只有a和b兩種活動，全能者必需即做a又做b。
不過，a和b不可能同時發生（至少表面看來確實如此），那麼邏輯上就不可能
有任何人事物能同時做到a和b。既然如此，X就不可能會是Y。除非梁提出
論證證立X就是Y，否則敗方仍然是梁，因為證明的責任(burden of proof)仍然
在梁那一方。
換句話說，問題重點不是有沒有能者可造出「沒有利器可刺破的盾牌」，及
有沒有能者可造出「可刺破任何盾牌的劍」，而是這兩人有沒有可能是同一個人。
所以李就算不用設想任何隱藏假設，梁的回答仍然未能觸及問題的核心。
Argument against omnipotence in a possible world with only a and b.
(1) X is omnipotent entails X can bring about any states of affairs. 
(Definition of omnipotence) 
(2) There are only two states of affairs to bring about in this possible world,
namely a and b. 
(Presumption) 
(3) X is omnipotent entails X bring about a. 
(From 1 and 2) 
(4) X is omnipotent entails X bring about b. (X is a dummy variable that we can
replace X with Y here.) 
(From 1 and 2) 
(5) X is omnipotent entails X can bring about a and b jointly. 
(From 1 and 2) 
(6) It is impossible for a and b to obtain jointly. 
(By definitions of a and b) 
(7) No X can bring about a and b jointly. 
(From 6) 
(8) No X can be omnipotent. 
(From 5 and 7) 
In Leung's discussion, it seems that he is not aware of the problem expressed by (6),
hence (7) and (8). My suggestion for a successful refutation is to modify the definition
of omnipotence expressed by (1). If omnipotence is so defined that any two states of
affairs in a possible world are not incompatible, i.e. there is no incompatible pairs of
states of affairs, (6) is false, hence so are (7) and (8).
