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Proof step 2 from proof uint2bits-sgn-2pot (previous step)

0 < j, ∀ i. ¬ i < 0 → j ≠ 2 ^ i ⊦ sgn(bitsinvert(uint2bits(j)) +1) = bitsinvert(uint2bits(j)) +1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
0 < i ⊦ ∃ bits. uint2bits(i) = false ' + true ' + bits
Proof step 4:
∃ bits. uint2bits(j) = false ' + true ' + bits, 0 < j, ∀ i. ¬ i < 0 → j ≠ 2 ^ i ⊦ sgn(bitsinvert(uint2bits(j)) +1) = bitsinvert(uint2bits(j)) +1

Rule argument

name: uint2bits-expos0 subst: [i] → [j]

Additional information

Used terms: j