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for example,we need show that if A and B are sets with A⊆B,then A⋃B.

1) $x \in A ⋃ B$

2) $(x \in A) \lor (x \in B)$

3) ∀$x(x \in A \to x \in B) $ --- by A⊆B given

4) $(x \in B) \lor (x \in B)$ ---we know that every x in A are also in B)

5) $(x \in B)$

6)$A ⋃ B⊆B$

Take other example, we have x which is domain consists of all people,

P(x) be "x is professor",Q(x) be "x is ignorant",R(x) be "x is in vain".

we have the proposition which are "No professors are are ignorant" and "All ignorant people are vain.".we have the followings:

1)∀$x(P(x) \to ¬Q(x)) $

2)∀$x(Q(x) \to R(x)) $

statement 1 is also equal to

∀$x(P(x) \to ¬(Q(x))) $

then we subtitute Q(x) by R(x),since statement 2 stated that every x in Q(x) is also in R(x).

finally we have,

∀$x(P(x) \to ¬(R(x))) $

which is incorrect.So my question why we cannot subtitute Q(x) by R(x)?

Moly Holy
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  • Because there is no reason to think that $Q$ and $R$ means the same ... And your counterexample is enough to prove that the substitution is not correct. – Mauro ALLEGRANZA Feb 21 '20 at 08:45
  • In order to perform a safe replacement of formulas you have to prove $\leftrightarrow$. See e.g Substitution of logical equivalents. – Mauro ALLEGRANZA Feb 21 '20 at 08:45
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    See also Counterexample to Replacement – Mauro ALLEGRANZA Feb 21 '20 at 08:50
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    The proof must be: from 2) and 3) apply Disjunction elim to 2): two cases: (a) $x \in B$ - it's Ok; (b) $x \in A$; by 3) we have $x \in B$. Thus, in both cases of disjunction: $x \in A \lor x \in B$ we have hat $x \in B$ holds. Thus, by Disjunction elim we conclude with $x \in B$ and by Conditional proof we have : $(x \in A \cup B) \to (x \in B)$. Then (as per your previous post), Generalize to get $\forall x [ \ldots]$, i.e. $(A \cup B) \subseteq B$. – Mauro ALLEGRANZA Feb 21 '20 at 09:21
  • I understand your method.What if i want to prove if A⊆B then P(A)⊆P(B)

    1.∀x(x∈A→x∈B) --- unwind definition of A⊆B

    2.x∈A

    3.{x}⊆A---subset that consist from x

    4.{x}∈P(A)

    5.x∈B ---from 1 and 2 using implication

    6.{x}⊆B

    7.{x}∈P(B)

    Before using Universal Generalization ,How can we construct condition proof,i mean {x}∈P(B)→{x}∈P(A) or {x}∈P(A)→{x}∈P(B)?I mean the rule of picking premise to construct conditiobal statement.

     – Moly Holy Feb 21 '20 at 10:02
  • Your goal should be $${\vdots\\forall C~(\forall x~(x\in C\to x\in A)~\to~\forall x~(x\in C\to x\in B))\\forall C~(C\subseteq A~\to~C\subseteq B)\\forall C~(C\in\mathcal P(A)\to C\in\mathcal P(B))\\mathcal P(A)\subseteq\mathcal P(B)}$$ – Graham Kemp Feb 21 '20 at 10:08
  • You have to start from an $x$ in $\mathcal P(A)$ and show that $x \in \mathcal P(B)$, using the fact that $A \subseteq B$. It is quite simple: if $x \in \mathcal P(A)$, then $x \subseteq A$. – Mauro ALLEGRANZA Feb 21 '20 at 10:09
  • @MauroALLEGRANZA 1).∀x(x∈A→x∈B) 2){x}∈P(A) 3){x}⊆A 4)x∈A 5)x∈B-- by using 1 and 4 6){x}⊆B 7){x}∈P(B) 8){x}∈P(A)→{x}∈P(B) 9)∀x({x}∈P(A)→{x}∈P(B)) by Universal Generalization because we are taking arbitrary element. 10)P(A)⊆P(B) .Is it like that?Or it can be unwind in deep. – Moly Holy Feb 21 '20 at 10:22
  • @GrahamKemp It is concise proof and good reference. – Moly Holy Feb 21 '20 at 10:22
  • @MolyHoly No, it does not suffice to show all singleton subsets of $A$ are subsets of $B$. It is necessary to show that any subset of $A$ is a subset of $B$. – Graham Kemp Feb 21 '20 at 10:35
  • @GrahamKemp if we use x to represent all the subset of A,then how can we represent the element from x? – Moly Holy Feb 21 '20 at 10:53
  • Use $C$ to represent an arbitrary subset of $A$. ($C\subseteq A$) $$\tiny\def\fch#1#2{~~\begin{array}{|l}#1\\hline#2\end{array}}\fch{A\subseteq B}{\forall x~(x\in A\to x\in B)\\fch{[C]:C\in\mathcal P(A)}{C\subseteq A\\forall x~(x\in C\to x\in A)\\fch{[x]:x\in C}{x\in A\to x\in B\x\in C\to x\in A\x\in A\x\in B}\\forall x~(x\in C\to x\in B)\C\subseteq B\C\in\mathcal P(B)}\\forall C~(C\in\mathcal P(A)\to C\in\mathcal P(B))\\mathcal P(A)\subseteq\mathcal P(B)}$$ – Graham Kemp Feb 21 '20 at 11:02

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So my question why we cannot subtitute Q(x) by R(x)?

Because $Q(x)$ and $R(x)$ have not been established to be equivalent. You can only substitute equivalents.

You have been promised that $\forall x~(Q(x)\to R(x))$, but you cannot assert from there that $\forall x~(R(x)\to Q(x))$; which means that you can not justify claiming that $\forall x~(\neg Q(x)\to\neg R(x))$.

Graham Kemp
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