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Proof step 83 from proof int2bits-add-high (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 2 flags: discard

Additional information

Used simplifier rules: ⊦ (boolvar ' + bits).rest = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ (boolvar ' + bits).first ↔ boolvar;