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Proof step 5 from proof int2bits-one-bit (previous step)

int2bits(i) = boolvar ', i < 0 ⊦ i = 0, i = ~1 ∧ boolvar

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ j < 0 ∧ j ≠ ~1 → int2bits(j) = int2bits((j - 1) / 2) + (~1 = j % 2)'

Rule argument

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

Additional information

Used simplifier rules: int2bits-mone-02: ⊦ int2bits(~1) = true ';%int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < j → ¬ j < i;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: i