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Proof step 13 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)) = bits2int(uint2bits(2 ^ n→i(# int2bits(~ j)) + j).rest)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
0 = 2 ^ n→i(# int2bits(~ j)) + j, j < 2 ^ n→i(# int2bits(~ j)), 0 < 2 ^ n→i(# int2bits(~ j)), j < 0, j + 1 < 2 ^ n→i(# int2bits(~ j)) ⊦ j = bits2int([])

Rule argument

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i;%uint2bits-zero: ⊦ uint2bits(0) = false ';% ⊦ i +1 = i + 1;% ⊦ i + j < j ↔ i < 0;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i < i + j ↔ 0 < j;% ⊦ boolvar '.rest = [];% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ 0 ≤ i ∧ i ≤ 1 → i = 0 ∨ i = 1; (forward)% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;