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单词 PropertiesOfSubstitutability
释义

properties of substitutability


In this entry, we list some basic properties of substitutability in first order logic with respect to commutativity.

Proposition 1.

If x,y are distinct variables, then

t[s/x][r/y]=t[r/y][s[r/y]/x],

provided that x does not occur in r.

Proof.

Suppose x and y are distinct variables, and r,s,t are terms. We do inductionMathworldPlanetmath on the complexity of t.

  1. 1.

    First, suppose t is a constant symbol. Then LHS and RHS are both t.

  2. 2.

    Next, suppose t is a variable.

    • If t is x, then LHS =s[r/y], and since y is not x, RHS =x[s[r/y]/x]=s[r/y].

    • If t is y, then LHS =y[r/y]=r, since x is not y, and RHS =r[s[r/y]/x]=r, since x does not occur in r.

    • If t is neither x nor y, then both sides are t.

    In all three cases, LHS = RHS.

  3. 3.

    Finally, suppose t is of the form f(t1,,tn). Then LHS =f(t1[s/x],,tn[s/x])[r/y]=f(t1[s/x][r/y],,tn[s/x][r/y]), which, by induction, is

    f(t1[r/y][s[r/y]/x],,tn[r/y][s[r/y]/x])

    or f(t1[r/y],,tn[r/y])[s[r/y]/x]=f(t1,,tn)[r/y][s[r/y]/x]=RHS.

Now, if s is y, then t[y/x][r/y]=t[r/y][r/x], and we record the following corollary:

Corollary 1.

If x is not in r and y not in t, then t[y/x][r/y]=t[r/x].

The only thing we need to show is the case when x is y, but this is also clear, as t[y/x][r/y]=t[x/x][r/x]=t[r/x].

With respect to formulasMathworldPlanetmathPlanetmath, we have a similar propositionPlanetmathPlanetmath:

Proposition 2.

If x,y are distinct variables, then

A[t/x][s/y]=A[s/y][t[s/y]/x],

provided that x does not occur in s, and t and s are respectively free for x and y in A.

Proof.

Suppose x and y are distinct variables, s,t terms, and A a wff. We do induction on the complexity of A.

  1. 1.

    First, suppose A is atomic.

    • A is of the form t1=t2, then LHS is t1[t/x][s/y]=t2[t/x][s/y] and we can apply the previous equation to both t1 and t2 to get RHS.

    • If A is of the form R(t1,,tn), then LHS is R(t1[t/x][s/y],,tn[t/x][s/y]), and we again apply the previous equation to each ti to get RHS.

  2. 2.

    Next, suppose A is of the form BC. Then LHS =B[t/x][s/y]C[t/x][s/y], and, by induction, is B[s/y][t[s/y]/x]C[s/y][t[s/y]/x]= RHS.

  3. 3.

    Finally, suppose A is of the form zB.

    • x is z. Then A[t/x][s/y]=A[s/y], and A[s/y][t[s/y]/x]=A[s/y] since x is bound in A[s/y].

    • x is not z. Then A[t/x][s/y]=(zB[t/x])[s/y].

      • *

        y is z. Then (zB[t/x])[s/y]=zB[t/x]. On the other hand, A[s/y][t[s/y]/x]=A[t[s/y]/x]=zB[t[s/y]/x]. By induction, t,s are free for x,y in B, and B[t/x]=B[t[s/y]/x], the result follows.

      • *

        y is not z. Then (zB[t/x])[s/y]=zB[t/x][s/y]. On the other hand, A[s/y][t[s/y]/x]=(zB[s/y])[t[s/y]/x]=zB[s/y][t[s/y]/x] since x is not z. By induction again, t,s are free for x,y in B, and B[t/x]=B[t[s/y]/x], the result follows once more.

Now, if t is y, then A[y/x][s/y]=A[s/y][s/x], and we record the following corollary:

Corollary 2.

If y is not free in A, and is free for x in A, then A[y/x][s/y]=A[s/x].

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