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Proof step 127 from proof int2bytes-rec-neg-05 (previous step)

j % 2 ^ (i * 8) = 0 ∨ j % 2 ^ (i * 8) = ~1, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), 0 ≤ i, j % 2 ^ (i * 8) < 0, j < 0, j + 2 ^ (i * 8) < i, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8 - 1) < 0, 0 < i ⊦ j % 2 ^ (i * 8) = 0, 2 ^ (i * 8) = 0, fix(i→n(i) + 1, int2bytes(j % 2 ^ (i * 8))).rest = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 128:
j % 2 ^ (i * 8) = ~1, 8 * i→n(i) = 0, 0 < i, j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i, j < 0, 0 ≤ i ⊦ 2 ^ (i * 8) = 0, fix(i→n(i) + 1, i2b(~1) ').rest = []

Rule argument

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ # boolvar ' = 0 +1;% ⊦ sgn(true ' + true ') = true ';% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + false ')+1 = bits + true;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bitsinvert(true ') = false ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;