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Proof step 66 from proof uint2bits-add-high (previous step)

uint2bits((j - 1) / 2 + 2 ^ (i - 1)) = sgn(int2bits((j - 1) / 2).rest), 2 ^ (i - 2) < 2 ^ i, 0 < 2 ^ i, 1 < 2 ^ i, 2 ^ (i - 1) + 1 < 2 ^ i, 0 < j + 2 ^ i, 2 ^ (i - 1) + j < 2 ^ i, j + 1 < i, 2 ^ (i - 1) + j < i, (2 ^ (i - 1) + j) + 1 < i, 2 ^ (i - 1) < 2 ^ i, 0 < i, 2 ^ (i - 1) + j < 0, j < i, j + 1 < 2 ^ i, j < 2 ^ i, (2 ^ (i - 1) + j) + 1 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ j + 2 ^ i, j ≤ 2 ^ i, j < 0, abs(j - 1) / 2 ≤ 2 ^ (i - 1), 2 ^ (i - 2) < abs(j - 1) / 2, abs(j - 1) / 2 < 2 ^ i, 2 ^ (i - 2) < 2 ^ (i - 1), ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ i = 1, j = 0, j = ~1, ~1 = j % 2, j + 2 ^ i = 1, uint2bits((j - 1) / 2 + 2 ^ (i - 1)) + false ' = sgn(int2bits((j - 1) / 2).rest + false '), true ∈ int2bits((j - 1) / 2).rest

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 67:
⊦ true ∈ int2bits(j).rest ↔ j ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ j = ~(2 ^ i))
Proof step 68:
uint2bits((j - 1) / 2 + 2 ^ (i - 1)) = sgn(int2bits((j - 1) / 2).rest), 2 ^ (i - 2) < 2 ^ i, 0 < 2 ^ i, 1 < 2 ^ i, 2 ^ (i - 1) + 1 < 2 ^ i, 0 < j + 2 ^ i, 2 ^ (i - 1) + j < 2 ^ i, j + 1 < i, 2 ^ (i - 1) + j < i, (2 ^ (i - 1) + j) + 1 < i, 2 ^ (i - 1) < 2 ^ i, 0 < i, 2 ^ (i - 1) + j < 0, j < i, j + 1 < 2 ^ i, j < 2 ^ i, (2 ^ (i - 1) + j) + 1 < 2 ^ i, 0 ≤ 2 ^ i, 0 ≤ j + 2 ^ i, j ≤ 2 ^ i, j < 0, abs(j - 1) / 2 ≤ 2 ^ (i - 1), 2 ^ (i - 2) < abs(j - 1) / 2, abs(j - 1) / 2 < 2 ^ i, 2 ^ (i - 2) < 2 ^ (i - 1), ∀ i0, j. i→n(i0) < i→n(i) → true → j < 0 ∧ 0 < i0 ∧ 2 ^ (i0 - 1) < abs(j) ∧ abs(j) ≤ 2 ^ i0 → uint2bits(j + 2 ^ i0) = sgn(int2bits(j).rest) ⊦ i = 1, j = 0, j = ~1, ~1 = j % 2, j + 2 ^ i = 1, uint2bits((j - 1) / 2 + 2 ^ (i - 1)) + false ' = sgn(int2bits((j - 1) / 2).rest + false '), (j - 1) / 2 ≠ 0 ∧ ¬(∃ i. 0 ≤ i ∧ (j - 1) / 2 = ~(2 ^ i))

Rule argument

name: int2bits-truein-01 rotate: F subst: [j] → [(j - 1) / 2]

Additional information

Used terms: (j - 1) / 2