Abstract
English translation of Ernst Schröder’s 1877 booklet Der Operationskreis des Logikkalkuls (Leipzig: Teubner) and his “Note über den Operationskreis des Logikcalculs,” Mathematische Annalen 12 (1877): 481–484. Schröder presents a reconstruction of Boole’s logical calculus that is “purified” in the sense that it admits only formulas readily interpretable as statements about relations between classes.
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Notes
- 1.
On this project and earlier related ones, see the abundant references in Trendelenburg (1867), pp. 1–63.
- 2.
[[W. S. Jevons (1864) is a striking omission from Schröder’s list.—Trans.]]
- 3.
- 4.
Boole’s decision to call 50) the “law of duality” seems wholly inappropriate in light of the real duality principle mentioned above and not fully recognized by Boole.
- 5.
[[Schröder might have added: every expression formed by addition and multiplication is equal to a product of class-symbols 0, 1, a, b, c, … and product-free sums. For example:
$$\displaystyle \begin{aligned}abc+cd=abc+dc=(ab+d)c=(a+d)(b+d)c\end{aligned}$$or
$$\displaystyle \begin{aligned}ab+ac+d=a(b+c)+d=(a+d)(b+c+d).\end{aligned}$$—Trans.]]
- 6.
[[Schröder’s three “postulates” are the existence and uniqueness of products, the existence and uniqueness of sums, and the existence and uniqueness of complements.—Trans.]]
- 7.
[[Translating the following more literally, we would get: “0 has itself in common with every other class” and “1 has in common with every other conceivable class that class itself.” Schröder does not actually use any word that translates as “members.”—Trans.]]
- 8.
- 9.
[[This seems rather an obscure way to say that Grassmann uses induction to prove, for example, the commutativity of addition. Grassmann has two units: e and − e. To prove that a + b = b + a, he first shows inductively that e + b = b + e and − e + b = b + −e. He then uses these facts in an inductive proof of the general result.—Trans.]]
- 10.
[[The German reads: “was auch bei 5) and 90) anginge.” What could this mean? There is no question of Grassmann offering inductive proofs of 5) and 90) or, indeed, any proofs at all since 90) is one of Grassmann’s axioms, while 50) and 5′) are not even theorems.—Trans.]]
- 11.
[[By 6′), addition of a and b to the first equation yields
$$\displaystyle \begin{aligned}a+ac=(a+b)(a+c)\end{aligned}$$$$\displaystyle \begin{aligned}(b+a)(b+c)=b+bc.\end{aligned}$$So, by 2′) and the second equation, a + ac = b + bc. Now apply 100.—Trans.]]
- 12.
[[That is, ∀a, b∃x, y(b = xa + ya 1 = (y + a)(x + a 1)).—Trans.]]
- 13.
[[That is, ∀a, b, u, v(b = (ab + ua 1)a + (a 1 b + va)a 1).—Trans.]]
- 14.
[[This seems to be a complicated way of saying that we can form instances of the equation
$$\displaystyle \begin{aligned}f(a)=xa+ya_1\end{aligned}$$by letting a be either 0 or 1. We then discover that
$$\displaystyle \begin{aligned}f(0)=x\cdot0+y\cdot(0)_1=x\cdot0+y\cdot1=y\end{aligned}$$$$\displaystyle \begin{aligned}f(1)=x\cdot1+y\cdot(1)_1=x\cdot1+y\cdot0=x.\end{aligned}$$—Trans.]]
- 15.
[[If f(a, b) = wab + xab 1 + ya 1 b + za 1 b 1, then f(1, 1) = w, f(1, 0) = x, etc.—Trans.]]
- 16.
[[So if no a’s are non-b’s and no b’s are non-a’s, a and b are the same class. This is the set-theoretic “axiom of extensionality”—though here it is a theorem, not an axiom.—Trans.]]
- 17.
[[We can now transform any expression into a product whose terms are all sums of class-symbols 0, 1, a, a 1, b, b 1, …. For example:
$$\displaystyle \begin{aligned}(a+b_1)_1+c+d=a_1b+c+d=(a_1+c)(b+c)+d=(a_1+c+d)(b+c+d).\end{aligned}$$It is easy to see that the equation
$$\displaystyle \begin{aligned}(a_1+c+d)(b+c+d)=1\end{aligned}$$is not a theorem since it is false when b = c = d = 0. Indeed, we now have a technique for determining whether any equation of the form α = 1 is a theorem: transform α into a product of sums, as above, and check whether any term of that product comes out 0 under an assignment of 0’s and 1’s to its summands. If there is such an assignment, the equation is not a theorem. If there is no such assignment, then each term must feature either 1 or a class-symbol together with its negation—in which case the equation is derivable from Schröder’s axioms. This is significant because Theorem 17′) says that every equation is equivalent to one of the form α = 1. So every equation true in every Boolean algebra is a theorem. (See Givant & Halmos 2009, ch. 15.) It is frustrating to see Schröder collect all the tools he needs for a completeness theorem and then pull up just short of the prize.—Trans.]]
- 18.
[[What final remark? (G) is a strong candidate.—Trans.]]
- 19.
[[For example:
$$\displaystyle \begin{aligned} (pab+ra_1b)_1=(pab+0\cdot ab_1+ra_1b+0\cdot a_1b_1)_1 \\ =p_1ab+1\cdot ab_1+r_1a_1b+1\cdot a_1b_1=p_1ab+ab_1+r_1a_1b+a_1b_1 \\ =(p_1a+r_1a_1)b+(a+a_1)b_1=(p_1a+r_1a_1)b+b_1 \\ =(p_1a+r_1a_1)b+(p_1a+r_1a_1)b_1+b_1=p_1a+r_1a_1+b_1. \end{aligned}$$As Schröder is just about to observe, there is a much easier way to apply 190) in this case:
$$\displaystyle \begin{aligned} (pab+ra_1b)_1=(pba+rba_1)_1=(pb)_1a+(rb)_1a_1 \\ =(p_1+b_1)a+(r_1+b_1)a_1=p_1a+r_1a_1+(a+a_1)b_1=p_1a+r_1a_1+b_1. \end{aligned}$$—Trans.]]
- 20.
[[That is, ∀x, y, a(xa + ya 1 = 0 ↔ (xy = 0 ∧∃u(a = ux 1 + y))). Note that xa + ya 1 is the shaded area in the following diagram.
The diagram makes it easy to see that, for example, xa + ya 1 = 0 only if xy = 0.—Trans.]]
- 21.
[[That is, if ∃u(a = ub 1), then ab = 0. So, as promised, we have an equivalence:
$$\displaystyle \begin{aligned}\forall a,b(ab=0\leftrightarrow\exists u(a=ub_1)).\end{aligned}$$Perhaps we would do a better job of highlighting the “arbitrariness” of u if we expressed the right-to-left conditional as
$$\displaystyle \begin{aligned}\forall a,b,u(a=ub_1\rightarrow ab=0).\end{aligned}$$All the talk about determination and arbitrariness seems an awkward way to discuss existential quantification. Schröder’s lemma provides a technique for eliminating existential quantifiers.—Trans.]]
- 22.
[[What Schröder may have in mind here is the inference
$$\displaystyle \begin{aligned}\exists u(ya_1=u_1y+xy)\Rightarrow\exists u(0=u_1y+xy)\end{aligned}$$in which our further “determination” of u is aided by our assumption that ya 1 = 0.—Trans.]]
- 23.
[[Schröder is struggling to describe the following inferences:
$$\displaystyle \begin{aligned}\exists u,v(a=ux_1\wedge u=v_1+y)\Rightarrow\exists v(a=(v_1+y)x_1)\Rightarrow\exists u(a=(u+y)x_1).\end{aligned}$$—Trans.]]
- 24.
[[That is, ∀a, x, y, u((a = ux 1 + y ∧ xy = 0) → (x 1 y 1 = 0 → a = y)).—Trans.]]
- 25.
[[Is Schröder saying that ∀a, x, y, u(a = ux 1 + y → (a = y → x 1 y 1 = 0))? This rendering does have the disadvantage of being false. If it were true, we would have ∀x, y, (y = yx 1 + y → x 1 y 1 = 0) from which it follows that ∀x, y(x 1 y 1 = 0). Another disadvantage of the proposed rendering is that it makes Schröder’s argument not just fallacious but patently so.—Trans.]]
- 26.
[[∀a, x, y, u((a = ux 1 + y ∧ xy = 0) → (x 1 y 1 = 0 ↔ a = y))? This rendering has the disadvantages mentioned in the previous footnote.—Trans.]]
- 27.
[[The given equation is equivalent to: (pb + qb 1)a + (rb + sb 1)a 1 = 0. If we eliminate a, the resultant is: prb + qsb 1 = 0. If we eliminate b from this equation, the resultant is, as promised, pqrs = 0.—Trans.]]
- 28.
[[xa + ya 1 + z = 0 if and only if xa + ya 1 = 0 and z = 0. If we eliminate a from xa + ya 1 = 0, the resultant is xy = 0. Adding back z = 0 we have xy + z = 0—which Schröder calls the “result” (Ergebniss) of eliminating a. This result gives a necessary condition for the truth of the original equation xa + ya 1 + z = 0. By the way, the result and the resultant are, as we shall see, the same in this case.—Trans.]]
- 29.
[[If we develop the equation pa + qa 1 + rb + sb 1 = 0, we get
$$\displaystyle \begin{aligned}(p+r)ab+(p+s)ab_1+(q+r)a_1b+(q+s)a_1b_1=0.\end{aligned}$$The product of the coefficients is pq + rs. So, if we eliminate a, b from the developed equation, the resultant is pq + rs = 0—the very “result” Schröder reaches via his shortcut.—Trans.]]
- 30.
[[The unified equation is: (x + p)a + (y + q)a 1 = 0.—Trans.]]
- 31.
[[First unify and develop:
$$\displaystyle \begin{aligned}0=xa+ya_1+z=xa+ya_1+z(a+a_1)=(x+z)a+(y+z)a_1.\end{aligned}$$Then the resultant is
$$\displaystyle \begin{aligned}0=(x+z)(y+z)=xy+xz+zy+z=xy+z.\end{aligned}$$—Trans.]]
- 32.
From Grassmann (1872). [[Grassmann’s Theorem 16 is Schröder’s 21′).—Trans.]]
- 33.
In Boole (1854), pp. 146–149. [[In the following statement of the problem, I have just copied down Boole’s language with some modifications that Schröder introduced in his German translation. Boole’s language, as modified, extends all the way down to the phrase “by which the given propositions will be expressed.”—Trans.]]
- 34.
[[It may be helpful to note that ab 1 = 0 (that is, a ⊆ b) if and only if ∃x(a = xb). As before, the business about “undetermined classes” is Schröder’s way of alerting us to an existential generalization.—Trans.]]
- 35.
[[In the case of α),
$$\displaystyle \begin{aligned}\exists x(a_1c_1=xe(bd_1+b_1d))\leftrightarrow a_1c_1(e(bd_1+b_1d))_1=0\leftrightarrow a_1c_1(e_1+bd+b_1d_1)=0.\end{aligned}$$As for γ), note that
$$\displaystyle \begin{aligned}a(b+e)(cd+c_1d_1)=a(b+e)(cd_1+c_1d)_1=(cd_1+c_1d)(cd_1+c_1d)_1=0\end{aligned}$$and
$$\displaystyle \begin{aligned}(cd_1+c_1d)(a_1+b_1e_1)=(cd_1+c_1d)(a(b+e))_1=a(b+e)(a(b+e))_1=0.\end{aligned}$$—Trans.]]
- 36.
[[Note that
$$\displaystyle \begin{aligned}cd+bc_1d_1=1cd+0cd_1+0c_1d+bc_1d_1\end{aligned}$$while
$$\displaystyle \begin{aligned}cd_1+c_1d+b_1c_1d_1=0cd+1cd_1+1c_1d+b_1c_1d_1.\end{aligned}$$—Trans.]]
- 37.
[[If we multiply cd 1 + c 1 d + b 1(c 1 + d 1) by c + c 1, we obtain cd 1 + c 1 d + b 1 c 1. Note, too, that
$$\displaystyle \begin{aligned}cd_1+c_1d+b_1c_1d_1=cd_1+c_1d+b_1c_1d+b_1c_1d_1=cd_1+c_1d+b_1c_1\end{aligned}$$and
$$\displaystyle \begin{aligned}cd_1+c_1d+b_1c_1d_1=cd_1+c_1d+b_1cd_1+b_1c_1d_1=cd_1+c_1d+b_1d_1.\end{aligned}$$—Trans.]]
- 38.
[[Note that
$$\displaystyle \begin{aligned} c+d=c(ad+ad_1+a_1d+a_1d_1)+d(ac+ac_1+a_1c+a_1c_1) \\ =acd+acd_1+ac_1d+a_1cd+a_1cd_1+a_1c_1d=acd_1+ac_1d+a_1cd. \end{aligned}$$using ζ) to delete acd, a 1 cd 1, and a 1 c 1 d.—Trans.]]
- 39.
[[δ) yields the homogeneous form:
$$\displaystyle \begin{aligned}(acd+ac_1d_1+a_1cd_1+a_1c_1d)b+(a_1c_1d_1)b_1=0.\end{aligned}$$Using 190) to negate the coefficient of b, we obtain:
$$\displaystyle \begin{aligned}(acd+ac_1d_1+a_1cd_1+a_1c_1d)_1=a_1cd+a_1c_1d_1+acd_1+ac_1d.\end{aligned}$$So, by 200),
$$\displaystyle \begin{aligned}b=(a_1cd+a_1c_1d_1+acd_1+ac_1d)v+a_1c_1d_1.\end{aligned}$$But a 1 c 1 d 1 v can be “incorporated in” a 1 c 1 d 1.—Trans.]]
- 40.
[[If you do the calculations, you will find that
$$\displaystyle \begin{aligned}w(c_1+d_1)+cd_1+c_1d=(w(c_1+d_1)+cd_1+c_1d)(c+c_1)(d+d_1)=wc_1d_1+cd_1+c_1d.\end{aligned}$$–Trans.]]
- 41.
[[The third premise immediately gives us: a 1 b 1 c = 0. The first two premises yield
$$\displaystyle \begin{aligned}ab(cd+c_1d_1)=0\;\;\;\text{and}\;\;\;bc(ad_1+a_1d)=0.\end{aligned}$$In particular, then,
$$\displaystyle \begin{aligned}abcd=0\;\;\;\text{and}\;\;\;abcd_1=0.\end{aligned}$$So
$$\displaystyle \begin{aligned}abc=abc(d+d_1)=abcd+abcd_1=0.\end{aligned}$$Since (bc)a + (b 1 c)a 1 = 0, 200) implies
$$\displaystyle \begin{aligned}a=u(bc)_1(b_1c)_1+b_1c=u(b_1+c_1)(b+c_1)+b_1c=uc_1+b_1c.\end{aligned}$$Similar reasoning confirms that b = vc 1 + a 1 c. Since (a 1 b 1 + ab)c = 0, 200) implies
$$\displaystyle \begin{aligned}c=w(a_1b_1+ab)_1=w(ab_1+a_1b).\end{aligned}$$—Trans.]]
- 42.
[[Cf. Schröder’s discussion of the two orders in Sect. 5.2.—Trans.]]
- 43.
[[By 170),
$$\displaystyle \begin{aligned}c+b=a\leftrightarrow(c+b)a_1+(c+b)_1a=0\leftrightarrow a_1c+a_1b+ab_1c_1=0.\end{aligned}$$By 200),
$$\displaystyle \begin{aligned}a_1c+ab_1c_1=0\leftrightarrow\exists u(c=ua+ab_1).\end{aligned}$$So
$$\displaystyle \begin{aligned}c+b=a\leftrightarrow(a_1b=0\wedge\exists u(c=ua+ab_1)).\end{aligned}$$By 170),
$$\displaystyle \begin{aligned}cb=a\leftrightarrow cba_1+(cb)_1a=0\leftrightarrow a_1bc+ac_1+ab_1=0.\end{aligned}$$By 200),
$$\displaystyle \begin{aligned}a_1bc+ac_1=0\leftrightarrow\exists u(c=u(a_1b)_1+a)\leftrightarrow\exists u(c=u(a+b_1)+a)\leftrightarrow\exists u(c=ub_1+a).\end{aligned}$$So
$$\displaystyle \begin{aligned}cb=a\leftrightarrow(ab_1=0\wedge\exists u(c=ub_1+a)).\end{aligned}$$–Trans.]]
- 44.
[[To reconcile 25′) with the reasoning in the previous note, consider the following. Assuming that a 1 b = 0,
$$\displaystyle \begin{aligned}b=(a+a_1)b=ab.\end{aligned}$$Furthermore,
$$\displaystyle \begin{aligned}b_1+u=b_1+u(b+b_1)=b_1+ub.\end{aligned}$$So
$$\displaystyle \begin{aligned}ab_1+ub=ab_1+uab=a(b_1+ub)=a(b_1+u)=ab_1+ua.\end{aligned}$$To make sense of 250), assume that ab 1 = 0. Then a = ab and, hence,
$$\displaystyle \begin{aligned}(a+b_1)(u+b)=ua +ab+ub_1=ua+a+ub_1=a+ub_1.\end{aligned}$$Note, finally, that b 1 = (a + a 1)b 1 = a 1 b 1.—Trans.]]
- 45.
[[If the four equations that follow are not already evident, you might note that:
$$\displaystyle \begin{aligned}a\div a=c\leftrightarrow c+a=a\leftrightarrow \exists u(c=ua)\end{aligned}$$$$\displaystyle \begin{aligned}a\div0=c\leftrightarrow c+0=a\leftrightarrow c=a\end{aligned}$$$$\displaystyle \begin{aligned}a::a=c\leftrightarrow ca=a\leftrightarrow \exists u(c=a+u)\end{aligned}$$$$\displaystyle \begin{aligned}a::1=c\leftrightarrow c\cdot1=a\leftrightarrow c=a.\end{aligned}$$–Trans.]]
- 46.
[[In each case, the “valence-condition” would be satisfied only if 0 = 1.—Trans.]]
- 47.
[[A lot is riding on the qualification “in essence” (im Grunde). For example, the equation
$$\displaystyle \begin{aligned}(1-(0+a))\cdot1=\frac{0}{1\cdot a}+0\end{aligned}$$is self-dual. Schröder would, presumably, say that this equation is essentially the same as 32). Cf. Schröder’s discussion of this question in the “Note” translated below.—Trans.]]
- 48.
[[By “an identity” (eine Identität) Schröder seems to mean an identity statement of the form α = α. The point is that the valence-condition for 31′) is 0 ⋅ b = 0—which “amounts to” 0 = 0 and is satisfied no matter what b is.—Trans.]]
- 49.
[[And so on? The point seems to be that similar remarks apply to 310) whose valence-condition, 0 ⋅ b 1 = 0, is satisfied no matter what b is.—Trans.]]
- 50.
[[Suppose c + b = a and cb = 0. Then
$$\displaystyle \begin{aligned}c=c(b+b_1)=cb+cb_1=cb_1=cb_1+bb_1=(c+b)b_1=ab_1=a-b.\end{aligned}$$Now suppose cb = a and c + b = 1. Then
$$\displaystyle \begin{aligned}c=c(b+b_1)=cb+cb_1=a+cb_1=a+cb_1+bb_1=a+(c+b)b_1=a+b_1=\frac{a}{b}.\end{aligned}$$—Trans.]]
- 51.
[[This is reminiscent of the idea that abstraction, as a psychological process, is a matter of selective attention. On this view, we derive our concept of cat from our experience of particular cats by disregarding or “abstracting from” any attributes not shared by all cats.—Trans.]]
- 52.
[[That is, propositions that include no more than three variables, all universally quantified.—Trans.]]
- 53.
[[If we combine all the valence-conditions for II and do some simplification, we get:
$$\displaystyle \begin{aligned}ac_1+a_1c+bc_1+b_1c=0.\end{aligned}$$But then, by 170), a = c and b = c and, hence,
$$\displaystyle \begin{aligned}a+(b-c)=a-(c-b)=(a-c)+b=b-(c-a).\end{aligned}$$None of these classes will equal (a + b) − c unless a, b, and c are all 0. All of this follows from the assumption that II as a whole is meaningful. If (as Schröder probably intends) we consider one equation at a time, we should note that the valence-conditions for, say, b − c and c − b imply that b = c. So the equation
$$\displaystyle \begin{aligned}a+(b-c)=a-(c-b)\end{aligned}$$will be true if it is even meaningful.—Trans.]]
- 54.
[[That is, an expression containing universally quantified variables.—Trans.]]
- 55.
[[(a + b) − b = (a + b)b 1 = ab 1 = a − b. The valence-condition for a − b is: a 1 b = 0. This yields: b = b(a + a 1) = ab. So (a + b) − b = a − ab. As for the other equations in I, first note that (a − b) + b = ab 1 + b = ab 1 + ab = a. The valence-condition for b − a is: ab 1 = 0. This yields: a = ab. So b − (b − a) = b − a 1 b = b(a 1 b)1 = b(a + b 1) = ab = a.—Trans.]]
- 56.
[[((a + b) − c) + ac = (a + b)c 1 + ac = (ac + ac 1) + bc 1 = a + bc 1 = a + (b − c). Note also: a + (b−c) = (ab + ab 1) + bc 1 = (a + c 1)b + ab 1 = (b−a 1 c) + ab 1 = (b− (c−a)) + ab 1.—Trans.]]
- 57.
[[(a+n)−(b+n) = (a+n)(b+n)1 = (a+n)b 1 n 1 = (ab 1)n 1 = (a−b)(1−n) = (a−b)−n(a−b).—Trans.]]
- 58.
[[Suppose, for example, that x + c = a + b. By 170),
$$\displaystyle \begin{aligned}(x+c)a_1b_1+x_1c_1(a+b)=0.\end{aligned}$$So a 1 b 1 c = 0 (which is the valence-condition for (a + b) ÷ c) and
$$\displaystyle \begin{aligned}xa_1b_1+x_1c_1(a+b)=0.\end{aligned}$$Applying 200) and 25′) to the latter equation, we have:
$$\displaystyle \begin{aligned}x=u(a+b)+c_1(a+b)=(a+b)(c_1+u)=(a+b)\div c.\end{aligned}$$—Trans.]]
- 59.
[[If c = au + bu then, ∃v, w(c = av + bw). On the other hand, suppose c = av + bw. Then
$$\displaystyle \begin{aligned}c=(av+avb)+(bw+abw)=(a+b)(av+bw).\end{aligned}$$So ∃u(c = (a + b)u = au + bu). We conclude:
$$\displaystyle \begin{aligned}\exists v,w(c=av+bw)\leftrightarrow\exists u(c=au+bu).\end{aligned}$$This seems to be what Schröder means when he says that av + bw = (a + b)u. Making sense of Schröder’s “arbitrary symbols” is not just a matter of prefixing existential quantifiers. When Schröder asserts that av + bw = (a + b)u, he clearly does not mean to be affirming the trivial fact that ∃u, v, w(av + bw = (a + b)u).—Trans.]]
- 60.
[[Given the first pair of assumptions, v 1 = a 1 b 1 u and, hence, abv 1 = 0. Furthermore,
$$\displaystyle \begin{aligned}a_1b+av=a_1b+a((a+b)u+u_1)=a_1b+(a+ab)u+au_1=a_1b+a(u+u_1)+abu=a_1b+a.\end{aligned}$$So
$$\displaystyle \begin{aligned}abv_1+(a_1b+av)w=(a_1b+a)(a+b)u=(a+a_1b+ab)u=(a+b)u.\end{aligned}$$On the other hand, given the last assumption,
$$\displaystyle \begin{aligned}(a+b)u=(a+b)(abv_1+(a_1b+av)w)=abv_1+a_1bw+avw+abvw=abv_1+(a_1b+av)w.\end{aligned}$$We conclude:
$$\displaystyle \begin{aligned}\exists v,w(c=abv_1+(a_1b+av)w)\leftrightarrow\exists u(c=(a+b)u).\end{aligned}$$—Trans.]]
- 61.
[[25′) and 250) provide some examples of such expressions.—Trans.]]
- 62.
[[Note that c + b = a only if a 1 b = 0. So if a 1 b ≠ 0, then (a − b) + b ≠ a.—Trans.]]
- 63.
[[Cf. 31′) and 310).—Trans.]]
- 64.
The theory of substitutions provides simple examples of mutually distributive operations that are neither commutative nor associative. Define a “symbolic” product and sum as follows:
$$\displaystyle \begin{aligned}a\cdot b=a^{\alpha}ba^{-\alpha},\;\;\;a+b=a^{\beta}b^{\gamma}a^{-\beta} .\end{aligned}$$Our notation distinguishes between the operation expressed by the multiplication-sign and the “genuine” multiplication of the substitutions expressed simply by placing the factors side by side. Note that the distributive law of arithmetic
$$\displaystyle \begin{aligned}a\cdot(b+c)=(a\cdot b)+(a\cdot c)\end{aligned}$$holds because both sides of this equation represent the expression a α b β c γ b −β a −α. Furthermore, the relation is mutual because we obtain a + (b ⋅ c) = (a + b) ⋅ (a + c) if we let γ = 1.
It is surprising that in the theory of substitutions, a field otherwise so well developed, distributive relations have attracted no attention.
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Pollard, S. (2022). The Operations of the Logical Calculus. In: Pollard, S. (eds) Ernst Schröder on Algebra and Logic . Synthese Library, vol 465. Springer, Cham. https://doi.org/10.1007/978-3-031-05671-0_5
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