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Proof step 2 from proof int2bits-inj (previous step)

int2bits(i) = int2bits(j) ⊦ i = j

User applied rule cut formula

And obtained the following new goals

Proof step 3:
int2bits(i) = int2bits(j) ⊦ bits2int(int2bits(i)) = bits2int(int2bits(j)), i = j

Rule argument

fma: bits2int(int2bits(i)) = bits2int(int2bits(j))

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i; Used formulas: bits2int(int2bits(i)) = bits2int(int2bits(j))