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Proof step 21 from proof int2int-neg (previous step)

0 < 2 ^ n→i(# int2bits(~ j)) + j, 0 ≤ 2 ^ n→i(# int2bits(~ j)) + j ⊦ j < 0 → bits2int(int2bits(j)) = 2 ^ n→i(# int2bits(~ j)) + j - 2 ^ (n→i(# uint2bits(2 ^ n→i(# int2bits(~ j)) + j)) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
j < 2 ^ n→i(# int2bits(~ j)), 1 < 2 ^ n→i(# int2bits(~ j)), j < 0, 0 < 2 ^ n→i(# int2bits(~ j)) + j, 0 ≤ 2 ^ n→i(# int2bits(~ j)) + j, 0 ≤ 2 ^ n→i(# int2bits(~ j)), 0 < 2 ^ n→i(# int2bits(~ j)), j + 1 < 2 ^ n→i(# int2bits(~ j)) ⊦ 2 ^ n→i(# int2bits(~ j)) = 2 ^ (n→i(# uint2bits(2 ^ n→i(# int2bits(~ j)) + j)) - 1)

Rule argument

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i +1 = i + 1;% ⊦ i < i + j ↔ 0 < j;% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i - j = i0 ↔ i = i0 + j;