Proof of Claim 3. By [c4] it suffices to prove the first equation. The proof is by induction on the cardinal k 1 + k 2 . The induction base k 1 = k 2 = 0 follows from sur 1 (s) = sur 2 (s) for df(s) = dc(s) = 0.

Suppose the claim is proved for all X k , Y k ′ with k ≤ m, k ′ ≤ r and k + k ′ < k 1 + k 2 . We prove the claim for X k 1 , Y k 2 . The proof is by induction on the construction of terms s with dF(s) < k 1 and d G (s) < k 2 . The boolean cases are trivial.

Suppose s = f(s 1 , . . , s t ) with d F (s) ≤ k 1 and d G (s) ≤ k 2 .  We have to show the following two statements:
(i) Xk, N s™,c = Xk, N sur1 (s)M,c
(ii) Y k 2 ∩ s M c  = Y k 2 ∩ s u r 1 (s) M c Consider (i) first. The induction hypothesis yields

$$X_{k\,_{1}-1}\,\cap\,s_{i}^{\mathfrak{M},\mathcal{C}}\,=\,X_{k\,_{1}-1}\,\cap\,s u r_{1}\left(s_{i}\right)^{\mathfrak{M},\mathcal{C}}$$
for $1\leq i\leq l$. We have:
$$X_{k_{1}-1}\cap c^{\mathfrak{M}}(X_{k_{1}-1})\cap s^{\mathfrak{M},\mathcal{C}}=X_{k_{1}-1}\cap c^{\mathfrak{M}}(X_{k_{1}-1})\cap f^{\mathfrak{M}}(s_{1}^{\mathfrak{M},\mathcal{C}},\ldots,s_{l}^{\mathfrak{M}})$$ $$=X_{k_{1}-1}\cap c^{\mathfrak{M}}(X_{k_{1}-1})\cap f^{\mathfrak{M}}(su_{1}(s_{1})^{\mathfrak{M},\mathcal{C}})$$ $$=X_{k_{1}-1}\cap c^{\mathfrak{M}}(X_{k_{1}-1})\cap sur_{1}(s)^{\mathfrak{M},\mathcal{C}}.$$
$${\mathcal{C}}_{\mathrm{)}}$$

The second equation is an immediate consequence of the third property of covering normal terms as given in Definition 26. Now the equation follows from [c5], i.e. X k, C X k,-1 N
c 20 (Xk, -1). (i) is proved.

(ii) Suppose first that k 2 = r. Then Y k 2 = X m and the claim can be proved as above since X m ⊆ X k 1 and, by induction hypothesis, X k 1 − 1 ∩ s M ′ C  = X k 1 − 1 ∩ sur1 (s i ) M ′ C 1 for 1 < : < 1.

Assume now that k 2 < r . By induction hypothesis,

$$\cap\;s_{i}^{\mathfrak{M},c}\,=\,Y_{k\;_{2}}\cap$$
$\square$
$s_i\,$)^\infty\,$. 
Yk, ∩ s™,c  = Yk, ∩ sur2 (si )™,c

$${\mathrm{for~}}1\leq i\leq l.{\mathrm{~Hence}}$$
$f^\circ$. 
f x ( Y k z ∩ s x ,c , . . ,Y k z ∩ s x ,c ) = f 20 ( Y k z ∩ sur 2 ( s 1 ) 20 ,c , . . ,Y k z ∩ sur 2 ( s t ) 20 ,c ).

We intersect both sides of the equation with Y k 2 and derive with the help of [c8]:
Yk 2 ∩ f 2 × ( s 1 ,c ,...,s 1 ,c ) = Yk 2 ∩ f 2 × ( sur 2 ( s 1 ) 2 × ,c ,...,sur 2 ( s t ) 2 ,c ).

This means Y k , N s 20 , C = Y k , N sur 2 (s) 20 , C , and the equation follows. The statements are proved.

The case s = g(s 1 , . . , s t ) is dual and left to the reader. We have proved claim 3.

By induction (c.f. in the proof of Theorem 20 the proof of (1) from the corresponding claim), we obtain from Claim 3:

$\mathbf{a}\cdot\mathbf{b}$
Xm N s M,c  = Xm N sur 1 ( s ) M,c  for all s E term ( Γ ).

(9)

$\sin x=\frac{\pi}{2}$. 
$$\mathbf{a}+\mathbf{b}=\mathbf{a}$$