let a, b be Element of INT ; :: thesis: for A, B being sequence of INT st A . 0 = a & B . 0 = b & ( for i being Nat holds
( A . (i + 1) = B . i & B . (i + 1) = (A . i) mod (B . i) ) ) holds
{ i where i is Nat : B . i = 0 } is non empty Subset of NAT
let A, B be sequence of INT; :: thesis: ( A . 0 = a & B . 0 = b & ( for i being Nat holds
( A . (i + 1) = B . i & B . (i + 1) = (A . i) mod (B . i) ) ) implies { i where i is Nat : B . i = 0 } is non empty Subset of NAT )
assume A1: ( A . 0 = a & B . 0 = b & ( for i being Nat holds
( A . (i + 1) = B . i & B . (i + 1) = (A . i) mod (B . i) ) ) ) ; :: thesis: { i where i is Nat : B . i = 0 } is non empty Subset of NAT
A2: for x being object st x in { i where i is Nat : B . i = 0 } holds
x in NAT

proof

let x be object ; :: thesis: ( x in { i where i is Nat : B . i = 0 } implies x in NAT )
assume x in { i where i is Nat : B . i = 0 } ; :: thesis: x in NAT
then ex i being Nat st
( x = i & B . i = 0 ) ;
hence x in NAT by ORDINAL1:def 12; :: thesis: verum

end;

ex m0 being Nat st B . m0 = 0

proof

assume A3: for m0 being Nat holds not B . m0 = 0 ; :: thesis: contradiction
A4: for m0 being Nat holds |.(B . m0).| <> 0 by A3, ABSVALUE:2;
A5: for i being Nat holds |.(B . (i + 1)).| <= |.(B . i).| - 1

proof

let i be Nat; :: thesis: |.(B . (i + 1)).| <= |.(B . i).| - 1
|.(B . (i + 1)).| = |.((A . i) mod (B . i)).| by A1;
then |.(B . (i + 1)).| < |.(B . i).| by A4, Th5;
then |.(B . (i + 1)).| + 1 <= |.(B . i).| by NAT_1:13;
then (|.(B . (i + 1)).| + 1) - 1 <= |.(B . i).| - 1 by XREAL_1:9;
hence |.(B . (i + 1)).| <= |.(B . i).| - 1 ; :: thesis: verum

end;

defpred S₁[ Nat] means |.(B . $1).| <= |.(B . 0).| - $1;
A6: S₁[ 0 ] ;
A7: for i being Nat st S₁[i] holds
S₁[i + 1]

proof

let i be Nat; :: thesis: ( S₁[i] implies S₁[i + 1] )
assume A8: S₁[i] ; :: thesis: S₁[i + 1]
A9: |.(B . (i + 1)).| <= |.(B . i).| - 1 by A5;
|.(B . i).| - 1 <= (|.(B . 0).| - i) - 1 by A8, XREAL_1:9;
hence S₁[i + 1] by A9, XXREAL_0:2; :: thesis: verum

end;

A10: for i being Nat holds S₁[i] from NAT_1:sch 2(A6, A7);
|.(B . |.(B . 0).|).| <= |.(B . 0).| - |.(B . 0).| by A10;
hence contradiction by A4, NAT_1:14; :: thesis: verum

end;

then consider m0 being Nat such that
A11: B . m0 = 0 ;
m0 in { i where i is Nat : B . i = 0 } by A11;
hence { i where i is Nat : B . i = 0 } is non empty Subset of NAT by A2, TARSKI:def 3; :: thesis: verum