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Proof step 11 from proof int2bits-ones (previous step)

j < 0 ⊦ bitsinvert(uint2bits(~ j)) +1 = [], j = ~1, false ∈ bitsinvert(uint2bits(~ j)) +1

User applied rule insert lemma

And obtained the following new goals

Proof step 12:
1 < j ⊦ ∃ bits. uint2bits(j) = false ' + true ' + bits ∧ bits ≠ []
Proof step 13:
∃ bits. uint2bits(~ j) = false ' + true ' + bits ∧ bits ≠ [], j < 0 ⊦ bitsinvert(uint2bits(~ j)) +1 = [], j = ~1, false ∈ bitsinvert(uint2bits(~ j)) +1

Rule argument

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

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: ~ j