Wlog E does not contain trivial equations t=t

Exercise 26: from Meseguer/Goguen

Let Σ =sig1,sorts N,E;functions0 :→N, s:N →N,1 :→E, f :N →E

Assumption: There is a specificationD = (Σ, E) with finite E, such thatT :=T_D ∼=A₁. Wlog E does not contain trivial equations t=t.

LetE =E1∪E2∪E3

1. E₁ contains only ground equations

2. Equations from E₂ match the pattern l[x] = r, l = r[x], l[x] = r[y], (where l[x] means that variablex occurs in the terml). I.e. no equation inE has the same variable at both sides.

3. Equations from E₃ match the pattern l[x] = r[x], i.e. at least one variable occurs on both sides.

We prove now, thatE₂=∅,E₃=∅, and E₁ must be infinite.

1. Claim: E2 =∅: Look at the cases l[x] =r,l=r[x], l[x] =r[y]

• l[x] =r∈E₂: ThenT_N =V_N∪ {sⁿ0|n∈N} ∪ {sⁿx|n∈N}andT_E =V_E∪ {1} ∪ {f t| t∈TN}. We have two cases: x∈VN orx∈VE:

a) x ∈V_N: Then l[x]≡sⁿx and r ≡s^m0, or l[x]≡f sⁿx and r ∈ {1, f s^m0}. But we clearly know, that A₁ is no model for the formulas ∀xsⁿx=s^m0,∀xf sⁿx= 1, and

∀xf sⁿx0f s^m0

b) x∈V_E: The l[x]≡x and r ∈ {1, f s^m0}. Again we clearly know that ∀xx= 1 and

∀xx=f s^m0 do not hold.

We conclude thatE2 does not contain equations of the form l[x] =r

• l=r[x]: analogous to the previous case

• l[x] =r[y]: Look at arbitrary ground terms substituted for y, we obtain r⁰ =l[x]. We know A₁ |=∀x, yl[x] =r[y] implies A₁ |=∀xl[x] =r⁰. The latter does not hold as seen in the first case.

We conclude thatE₂ does not contain equations of the form l[x] =r[y]

Together we knowE2 =∅

2. Similarly we prove that E3 is empty. The main problem is to identify the cases, where a variable may occur twice.

3. Now toE1: Letl=r ∈E1, thenl≡sⁿ0 andr≡s^m0 or l∈ {f sⁿ0,1} andr ∈ {f s^m0,1}.

IfA₁|=sⁿ0 =s^m0 then n=m and we have a trivial equation.

If A₁ |= f sⁿ0 =f s^m0 then only for n 6=m we have a non-trivial equation, with both n, m either even or odd.

IfA₁|=f sⁿ0 = 1 then n must be odd.

Now assumeE₁ to be finite. ThenE₁ ={f sⁿⁱ =f s^mi |i= 1, . . . k} ∪ {f s^li = 1|i= 1, . . . k⁰}. Let q= 1 + max({l_i} ∪ {m_i} ∪ {n_i}). Then we know E10f s^q0 = 1 andE1 0f s^q0 =s^q0. SinceE2 =∅ andE₃=∅ we know A₁T for finite E₁.

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