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Proof step 77 from proof int2bits-add-high (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 78:
⊦ bits ≠ [] → (fillsgn(n, bits).first ↔ bits.first)
Proof step 79:
int2bits(j0).rest ≠ [], 2 ^ n→i(# int2bits(1 - j)) + j = 2 ^ i0, 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), ∃ i0. 1 = 2 ^ i0 + j

Rule argument

seq: ⊦ bits ≠ [] → (fillsgn(n, bits).first ↔ bits.first) subst: [bits, n] → [int2bits(j0).rest, i→n(i)] rotate: F spec: bitopsaux inst: name: fill-first

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ fillsgn(n, []) = mklist(false, n);% ⊦ sgn(true ' + mklist(false, n)) = true ' + mklist(false, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1; Used terms: int2bits(j0).rest, i→n(i)