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Proof step 7 from proof int2bits-restsgn (previous step)

0 ≤ 2 ^ n→i(# int2bits(j)) - j, 0 ≤ j ⊦ sgn(uint2bits(2 ^ n→i(# int2bits(j)) - j).rest) = uint2bits(2 ^ n→i(# int2bits(j)) - j).rest, ∃ i. 2 ^ i = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ j, j ≤ 2 ^ n→i(# int2bits(j)), 0 ≤ 2 ^ n→i(# int2bits(j)) ⊦ sgn(uint2bits(2 ^ n→i(# int2bits(j)) - j).rest) = uint2bits(2 ^ n→i(# int2bits(j)) - j).rest, ∃ i. 2 ^ i = j

Rule argument

Additional information

Used simplifier rules: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;