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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ i → int2bits(i) = uint2bits(i)
Proof step 4:
¬ 0 ≤ 2 ^ n→i(# int2bits(j)) - j, 0 ≤ j ⊦ sgn(int2bits(2 ^ n→i(# int2bits(j)) - j).rest) = int2bits(2 ^ n→i(# int2bits(j)) - j).rest, ∃ i. 2 ^ i = j
Proof step 7:
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

Rule argument

name: int2bits-pos rotate: F subst: [i] → [2 ^ n→i(# int2bits(j)) - j]

Additional information

Used terms: 2 ^ n→i(# int2bits(j)) - j