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Proof step 15 from proof int2bits-truein-01 (previous step)

¬ true ∈ bits, bits2int(int2bits(j)) = ~(2 ^ n→i(# bits)), int2bits(j) = true ' + bits, j < 0 ⊦ j = 0, true ∈ bits, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
int2bits(j) = true ' + bits, 2 ^ n→i(# bits) + j = 0, j < 2 ^ n→i(# bits), 0 < 2 ^ n→i(# bits), j < 0, j + 1 < 2 ^ n→i(# bits) ⊦ j = 0, true ∈ bits, ∃ i. ¬ i < 0 ∧ 2 ^ i + j = 0

Rule argument

Additional information

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