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Proof step 5 from proof int2bits-exneg (previous step)

i < ~1, i < 0 ⊦ ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

User applied rule insert lemma

And obtained the following new goals

Proof step 6:
1 < j ⊦ ∃ bits. uint2bits(j) = false ' + true ' + bits ∧ bits ≠ []
Proof step 7:
∃ bits. uint2bits(~ i) = false ' + true ' + bits ∧ bits ≠ [], i < ~1, i < 0 ⊦ ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

Rule argument

name: uint2bits-expos1 subst: [j] → [~ i]

Additional information

Used simplifier rules: ⊦ i < ~ j ↔ i + j < 0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: ~ i