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Proof step 5 from proof int2uint-fill (previous step)

0 ≤ j, fillsgn(n, int2bits(j)).first ⊦ bits2uint(fillsgn(n, int2bits(j))) = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ bits ≠ [] → (fillsgn(n, bits).first ↔ bits.first)

Rule argument

seq: ⊦ bits ≠ [] → (fillsgn(n, bits).first ↔ bits.first) subst: [bits, n] → [int2bits(j), n] rotate: F spec: bitopsaux inst: name: fill-first

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: int2bits(j), n