reserve i for Nat, x,y for set;
reserve S for non empty non void ManySortedSign;
reserve X for non-empty ManySortedSet of S;

theorem Th62:
  for n,m being Nat, s,r being set
  for S1,S2 being non empty ConnectivesSignature st
    1 <= n & len the connectives of S1 >= n+3 &
    for i st i >= n & i <= n+3 holds
    (the Arity of S1).((the connectives of S1).i)
    = (the Arity of S2).((the connectives of S2).(i+m)) &
    (the ResultSort of S1).((the connectives of S1).i)
    = (the ResultSort of S2).((the connectives of S2).(i+m))
  holds S2 is (n+m,s,r)-array implies S1 is (n,s,r)-array
  proof
    let n,m be Nat;
    let s,r be set;
    let S1,S2 be non empty ConnectivesSignature;
    assume A1: 1 <= n;
    assume A2: len the connectives of S1 >= n+3;
    assume A3: for i st i >= n & i <= n+3 holds
    (the Arity of S1).((the connectives of S1).i)
    = (the Arity of S2).((the connectives of S2).(i+m)) &
    (the ResultSort of S1).((the connectives of S1).i)
    = (the ResultSort of S2).((the connectives of S2).(i+m));
    assume len the connectives of S2 >= n+m+3;
    given J,K,L being Element of S2 such that
A4: L = s & K = r & J <> L & J <> K &
    (the connectives of S2).(n+m) is_of_type <*J,K*>, L &
    (the connectives of S2).(n+m+1) is_of_type <*J,K,L*>, J &
    (the connectives of S2).(n+m+2) is_of_type <*J*>, K &
    (the connectives of S2).(n+m+3) is_of_type <*K,L*>, J;
    thus len the connectives of S1 >= n+3 by A2;
A5: n <= n+3 by NAT_1:11;
    then (the Arity of S1).((the connectives of S1).n)
    = (the Arity of S2).((the connectives of S2).(n+m)) &
    (the ResultSort of S1).((the connectives of S1).n)
    = (the ResultSort of S2).((the connectives of S2).(n+m)) by A3;
    then
A6: (the Arity of S1).((the connectives of S1).n) = <*J,K*> &
    (the ResultSort of S1).((the connectives of S1).n) = L by A4;
A7: n <= n+1 & n+1 <= n+1+2 & n+3 = n+1+2 by NAT_1:11;
    then (the Arity of S1).((the connectives of S1).(n+1))
    = (the Arity of S2).((the connectives of S2).(n+1+m)) &
    (the ResultSort of S1).((the connectives of S1).(n+1))
    = (the ResultSort of S2).((the connectives of S2).(n+1+m)) by A3;
    then
A8: (the Arity of S1).((the connectives of S1).(n+1)) = <*J,K,L*> &
    (the ResultSort of S1).((the connectives of S1).(n+1)) = J by A4;
A9: n+2 <= n+2+1 & n <= n+2 by NAT_1:11;
    then (the Arity of S1).((the connectives of S1).(n+2))
    = (the Arity of S2).((the connectives of S2).(n+2+m)) &
    (the ResultSort of S1).((the connectives of S1).(n+2))
    = (the ResultSort of S2).((the connectives of S2).(n+2+m)) by A3;
    then
A10: (the Arity of S1).((the connectives of S1).(n+2)) = <*J*> &
    (the ResultSort of S1).((the connectives of S1).(n+2)) = K by A4;
    n <= n+3 by NAT_1:11;
    then (the Arity of S1).((the connectives of S1).(n+3))
    = (the Arity of S2).((the connectives of S2).(n+3+m)) &
    (the ResultSort of S1).((the connectives of S1).(n+3))
    = (the ResultSort of S2).((the connectives of S2).(n+3+m)) by A3;
    then
A11: (the Arity of S1).((the connectives of S1).(n+3)) = <*K,L*> &
    (the ResultSort of S1).((the connectives of S1).(n+3)) = J by A4;
    n <= len the connectives of S1 & 1 <= n+1 & 1 <= n+1+1 &
    n+1 <= len the connectives of S1 & n+2 <= len the connectives of S1 &
    1 <= n+3 by A2,A5,A7,A9,XXREAL_0:2,NAT_1:11;
    then n in dom the connectives of S1 & n+1 in dom the connectives of S1 &
    n+2 in dom the connectives of S1 by A1,FINSEQ_3:25;
    then reconsider J,K,L as Element of S1 by A6,A8,A10,FUNCT_1:102,FUNCT_2:5;
    take J,K,L;
    thus L = s & K = r & J <> L & J <> K by A4;
    thus thesis by A6,A8,A10,A11;
  end;
