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Proof step 2 from proof int2bits-rec-neg-03 (previous step)

j < ~1, j < 0 ⊦ int2bits(j) = int2bits(j / 2 - 1) + true ', j % 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ j < 0 ∧ j ≠ ~1 → int2bits(j) = int2bits((j - 1) / 2) + (~1 = j % 2)'
Proof step 4:
j < 0 ∧ j ≠ ~1, j < ~1, j < 0 ⊦ int2bits((j - 1) / 2) + (~1 = j % 2)' = int2bits(j / 2 - 1) + true ', j % 2 = 0

Rule argument

name: int2bits-rec-neg rotate: F subst: [j] → [j]

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 ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ 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 ';% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ boolvar ' ≠ []; Used terms: j