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Proof step 6 from proof uint2bits-restsgn (previous step)

j < 2 * 2 ^ i, 2 ^ i < j, 0 < i, 2 < j, 0 ≤ 2 ^ i, 1 < j, 0 ≤ j, 0 < j, 1 < 2 ^ i, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ sgn(uint2bits(2 ^ (i + 2) - j).rest) = uint2bits(2 ^ (i + 2) - j).rest, ¬ 2 ^ (i + 2) ≤ j ∧ 2 ^ (i + 2) ≠ 1 + j

User applied rule case distinction

And obtained the following new goals

Proof step 7:
j < 2 * 2 ^ i, 2 ^ i < j, 0 < i, 2 < j, 0 ≤ 2 ^ i, 1 < j, 0 ≤ j, 0 < j, 1 < 2 ^ i, 0 < 2 ^ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 < i ∧ 0 ≤ j0 ∧ 2 ^ i < j0 ∧ j0 < 2 * 2 ^ i → sgn(uint2bits(2 ^ (i + 2) - j0).rest) = uint2bits(2 ^ (i + 2) - j0).rest ⊦ ¬ 2 ^ (i + 2) ≤ j, sgn(uint2bits(2 ^ (i + 2) - j).rest) = uint2bits(2 ^ (i + 2) - j).rest

Rule argument

pos: right 2

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ i + j = j + i;% ⊦ i + j - i = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ (boolvar ' + bits).rest = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ sgn(boolvar ') = boolvar ';% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;