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Proof step 46 from proof int2bits-neg-03 (previous step)

0 = 2 ^ n→i(# int2bits(~ j)) + j, 0 < 2 ^ n→i(# int2bits(~ j)), j < 0, j < 2 ^ n→i(# int2bits(~ j)), j + 1 < 2 ^ n→i(# int2bits(~ j)) ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 47:
0 ≤ j ⊦ j < 2 ^ (n→i(# int2bits(j)) - 1)
Proof step 48:
~ j < 2 ^ (n→i(# int2bits(~ j)) - 1), 0 = 2 ^ n→i(# int2bits(~ j)) + j, 0 < 2 ^ n→i(# int2bits(~ j)), j < 0, j < 2 ^ n→i(# int2bits(~ j)), j + 1 < 2 ^ n→i(# int2bits(~ j)) ⊦

Rule argument

name: int2bits-less-2pot subst: [j] → [~ j]

Additional information

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