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Proof step 25 from proof uint2bits-falsein-01 (previous step)

2 ^ i = 2 * (i0 + 1), 0 < i0, false ∈ uint2bits(i0).rest, i ≤ 0, ∀ i. 2 ^ i ≠ i0 + 1, ∀ j0. i→n(j0) < i→n(i0 * 2 + 1) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ 2 ^ (i - 1) = i0 + 1, i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ i < 0 → 2 ^ i = 0

Rule argument

seq: ⊦ i < 0 → 2 ^ i = 0 subst: [i] → [i] rotate: F spec: int-pot inst: name: two-minus

Additional information

Used simplifier rules: ⊦ 0 < i ∧ 0 < j → 0 ≠ i + j;% ⊦ 2 * i = 0 ↔ i = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i; Used terms: i