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Proof step 11 from proof bint-fillsgn-len (previous step)

32 < # int2bits(i), ~1 ≤ i, i ≤ 0 ⊦ # fillsgn(32, int2bits(i)) = 32

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ i + 1 = 0 → int2bits(i) = true ' subst: [i] → [i] rotate: F spec: int2bits inst: name: int2bits-mone

Additional information

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