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Proof step 10 from proof uint2bits-add-high-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 11:
# int2bits(j0) ≤ i→n(i) + 1, 1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ i ⊦ fillsgn(i→n(i) + 1, sgn(int2bits(j0).rest)).rest = fillsgn(i→n(i), int2bits(j0).rest)
Proof step 13:
1 < j, 0 < j, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 ≤ j0 + 2 ^ i, 0 ≤ i ⊦ # int2bits(j0) ≤ i→n(i) + 1, fillsgn(i→n(i) + 1, sgn(int2bits(j0).rest)).rest = fillsgn(i→n(i), int2bits(j0).rest)

Rule argument

fma: # int2bits(j0) ≤ i→n(i) + 1

Additional information

Used formulas: # int2bits(j0) ≤ i→n(i) + 1