Are the BSIG0, BSIG1, SSIG0 (x), SSIG0 (x) functions reversible in the SHA-256 algorithm?

BSIG0(x) = ROTR^2(x) XOR ROTR^13(x) XOR ROTR^22(x) BSIG1(x) = ROTR^6(x) XOR ROTR^11(x) XOR ROTR^25(x) SSIG0(x) = ROTR^7(x) XOR ROTR^18(x) XOR SHR^3(x) SSIG1(x) = ROTR^17(x) XOR ROTR^19(x) XOR SHR^10(x)

That is, for example:

For the function y=BSIG0(x) , is it possible to find a function F such that the condition:

 x=F(BSIG0(x))
  • What kind of function in question is not clear. In addition, there is a suspicion that the author of the question did not want to read the articles on SHA256. Therefore, under the closure. - Vladimir Martyanov
  • Functions S0, S1, E0, E1. - Denis
  • In addition, no document says whether they are reversible or not. Or do you have other documentation? - Denis
  • The answer in the understanding of the abbreviation SHA - Yevgeny Borisov
  • SHA-2 (English Secure Hash Algorithm Version 2 - secure hashing algorithm, version 2). And where is it about the reversibility of these functions? - Denis

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