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Proof step 2 from proof int2bits-rec-pos-01 (previous step)

1 < i, 0 < i ⊦ int2bits(i) = int2bits(i / 2) + (i % 2 = 1)'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ 1 < i → int2bits(i) = int2bits(i / 2) + (i % 2 = 1)

Rule argument

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

Additional information

Used simplifier rules: ⊦ bits + boolvar = bits + boolvar ';% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used terms: i