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Proof step 7 from proof int2uint-fill (previous step)

¬ fillsgn(n, int2bits(j)).first ∧ fillsgn(n, int2bits(j)) ≠ [], 0 ≤ j ⊦ bits2int(fillsgn(n, int2bits(j))) = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ bits ≠ [] → bits2int(fillsgn(n, bits)) = bits2int(bits)

Rule argument

name: bits2int-fill rotate: F subst: [bits, n] → [int2bits(j), n]

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2int-inj: ⊦ bits2int(int2bits(i)) = i;%int2bits-not-empty: ⊦ int2bits(j) ≠ []; Used terms: int2bits(j), n