reconsider y = x mod p as Element of INT by INT_1:def 2;
then consider s being Element of INT such that
A5:
( ( (ALGO_EXGCD (p,y)) `3_3 = 1 implies ( ( (ALGO_EXGCD (p,y)) `2_3 < 0 implies ex z being Element of INT st
( z = (ALGO_EXGCD (p,y)) `2_3 & s = p + z ) ) & ( 0 <= (ALGO_EXGCD (p,y)) `2_3 implies s = (ALGO_EXGCD (p,y)) `2_3 ) ) ) & ( (ALGO_EXGCD (p,y)) `3_3 <> 1 implies s = {} ) )
;
take
s
; for y being Element of INT st y = x mod p holds
( ( (ALGO_EXGCD (p,y)) `3_3 = 1 implies ( ( (ALGO_EXGCD (p,y)) `2_3 < 0 implies ex z being Element of INT st
( z = (ALGO_EXGCD (p,y)) `2_3 & s = p + z ) ) & ( 0 <= (ALGO_EXGCD (p,y)) `2_3 implies s = (ALGO_EXGCD (p,y)) `2_3 ) ) ) & ( (ALGO_EXGCD (p,y)) `3_3 <> 1 implies s = {} ) )
thus
for y being Element of INT st y = x mod p holds
( ( (ALGO_EXGCD (p,y)) `3_3 = 1 implies ( ( (ALGO_EXGCD (p,y)) `2_3 < 0 implies ex z being Element of INT st
( z = (ALGO_EXGCD (p,y)) `2_3 & s = p + z ) ) & ( 0 <= (ALGO_EXGCD (p,y)) `2_3 implies s = (ALGO_EXGCD (p,y)) `2_3 ) ) ) & ( (ALGO_EXGCD (p,y)) `3_3 <> 1 implies s = {} ) )
by A5; verum