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Proof step 8 from proof int2bits-len-pos-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ j < 0 → abs(j) = 0 - j

Rule argument

seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i + 1] rotate: F spec: int-abs inst: name: Abs-neg

Additional information

Used simplifier rules: int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0;%int2bits-eq: ⊦ int2bits(j) = false ' ↔ j = 0;% ⊦ 0 - i = ~ i;% ⊦ i < ~ j ↔ i + j < 0;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i < 0 → abs(i) = ~ i;% ⊦ ~ i + i = 0;% ⊦ ¬ j < 0 → abs(j) = j;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: i + 1