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Proof step 77 from proof shiftright-rec-neg (previous step)

int2bits(i % 2).butlast = [], i < ~1, i < 0 ⊦ abs(j) = 1, i→n(j) + # int2bits(i % 2) = 0, int2bits((i - 1) / 2) + int2bits(i % 2) = [], j = 0, bits2int(true ' + firstn(# int2bits((i - 1) / 2), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2))) = bits2int(firstn(# int2bits((i - 1) / 2), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2))), mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2) = [], (mklist(true, i→n(j) - 1) + int2bits((i - 1) / 2)).first

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ n ≠ 0 → ((mklist(boolvar, n) + bits).first ↔ boolvar) subst: [bits, n, boolvar] → [int2bits((i - 1) / 2), i→n(j) - 1, true] rotate: F spec: list-last inst: bitlista name: mk-first

Additional information

Used simplifier rules: ⊦ mklist(boolvar, 0) = [];% ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ int2bits(j) ≠ [];% ⊦ firstn(# bits, bits) = bits;% ⊦ bits2int(int2bits(i)) = i;% ⊦ int2bits(j).first ↔ j < 0;%0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false); Used terms: int2bits((i - 1) / 2), i→n(j) - 1, true