reserve i,n for Nat;
reserve m for non zero Nat;
reserve p,q for Tuple of n, BOOLEAN;
reserve d,d1,d2 for Element of BOOLEAN;

theorem
  for z1 being Tuple of 2, BOOLEAN holds
  z1=<*FALSE*>^<*TRUE*> implies Intval(z1) = -2
proof
  let z1 be Tuple of 2,BOOLEAN;
  assume
A1: z1 = <*FALSE*>^<*TRUE*>;
  consider k1,k2 being Element of NAT such that
A2: Binary (z1) = <* k1,k2 *> by FINSEQ_2:100;
A3: z1 = <*FALSE,TRUE*> by A1,FINSEQ_1:def 9;
then A4: z1/.1 = FALSE by FINSEQ_4:17;
A5: z1/.2 = TRUE by A3,FINSEQ_4:17;
then A6: Intval(z1) = Absval(z1) - 2 to_power (1 + 1) by Def3
    .= Absval(z1) - (2 to_power 1 * 2 to_power 1) by POWER:27
    .= Absval(z1) - (2 * 2 to_power 1) by POWER:25
    .= Absval(z1) - (2 * 2) by POWER:25
    .= Absval(z1) - 4;
 1 in Seg 1 & Seg 1 c= Seg 2 by FINSEQ_1:3,5;
then
A7: (Binary(z1))/.1 = IFEQ(z1/.1,FALSE,0,2 to_power(1-'1)) by BINARITH:def 3
    .= 0 by A4,FUNCOP_1:def 8;
 2 in Seg 2 by FINSEQ_1:3;
then A8: (Binary(z1))/.2 = IFEQ(z1/.2,FALSE,0,2 to_power(2-'1)) by
BINARITH:def 3
    .= 2 to_power(2-'1) by A5,FUNCOP_1:def 8;
 2 - 1 = 1;
then  2 -' 1 = 1 by XREAL_0:def 2;
then A9: (Binary(z1))/.2 = 2 by A8,POWER:25;
   (Binary(z1))/.1 = k1 & (Binary(z1))/.2 = k2 by A2,FINSEQ_4:17;
then  Absval(z1) = addnat $$ (<* 0,2 *>) by A2,A7,A9,BINARITH:def 4
    .= addnat $$ (<* 0 *>^<* 2 *>) by FINSEQ_1:def 9
    .= addnat.(addnat$$<* 0 *>,addnat$$<* 2 *>) by FINSOP_1:5
    .= addnat.(0,addnat$$<* 2 *>) by FINSOP_1:11
    .= addnat.(0,2) by FINSOP_1:11
    .= 0 + 2 by BINOP_2:def 23
    .= 2;
  hence thesis by A6;
end;
