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Proof step 3 from proof int2int-neg (previous step)

¬ 0 ≤ 2 ^ n→i(# int2bits(~ j)) + j ⊦ j < 0 → bits2int(int2bits(j)) = bits2int(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
j < 0, 2 ^ n→i(# int2bits(~ j)) + j < 0 ⊦ j = bits2int(int2bits(2 ^ n→i(# int2bits(~ j)) + j).rest)

Rule argument

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i;% ⊦ ¬ i ≤ j ↔ j < i;