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Proof step 80 from proof int2bits-add-high (previous step)

2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, int2bits(j0).rest.first, 0 ≤ 2 ^ i * 2 ^ i0 + j0, 0 ≤ j0 + 2 ^ i, j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ i, j * 2 ^ i + j0 < 0, 0 < i0 ⊦ int2bits(j0).rest = [], sgn(true ' + fillsgn(i→n(i), int2bits(j0).rest)) = true ' + fillsgn(i→n(i), int2bits(j0).rest), i0 = 0, ∃ i0. 1 = 2 ^ i0 + j

User applied rule insert lemma

And obtained the following new goals

Rule argument

name: int2bits-exneg subst: [i] → [j0]

Additional information

Used simplifier rules: int2bits-neg-num: i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);%uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ sgn(true ' + true ') = true ';% ⊦ boolvar '.rest = [];% ⊦ fillsgn(n, []) = mklist(false, n);% ⊦ sgn(true ' + mklist(false, n)) = true ' + mklist(false, n);% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1; Used terms: j0