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Proof step 16 from proof int2bits-add-high (previous step)

j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0 < 0 ⊦ sgn(int2bits(2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

User applied rule cut formula

And obtained the following new goals

Proof step 17:
abs(j) < 2 ^ n→i(# int2bits(~ j)), j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0 < 0 ⊦ sgn(int2bits(2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)
Proof step 22:
j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0 < 0 ⊦ abs(j) < 2 ^ n→i(# int2bits(~ j)), sgn(int2bits(2 ^ i * (2 ^ n→i(# int2bits(~ j)) + j) + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

fma: abs(j) < 2 ^ n→i(# int2bits(~ j))

Additional information

Used formulas: abs(j) < 2 ^ n→i(# int2bits(~ j))