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Proof step 25 from proof int2int-neg (previous step)

j < i, i < n→i(# int2bits(~ j)), 2 ^ i + j < 2 ^ n→i(# int2bits(~ j)), 0 < 2 ^ i, i ≤ n→i(# int2bits(~ j)), 2 ^ i + j ≤ 2 ^ n→i(# int2bits(~ j)), 2 ^ i + j ≤ i, j < 2 ^ n→i(# int2bits(~ j)), 1 < 2 ^ n→i(# int2bits(~ j)), 0 < 2 ^ n→i(# int2bits(~ j)) + j, 0 ≤ 2 ^ n→i(# int2bits(~ j)) + j, 0 ≤ 2 ^ n→i(# int2bits(~ j)), 0 < 2 ^ n→i(# int2bits(~ j)), j + 1 < 2 ^ n→i(# int2bits(~ j)), 2 ^ i + j ≤ 0, 0 ≤ i, 0 < 2 ^ i * 2 + j, j < n→i(# int2bits(~ j)), j + 1 < n→i(# int2bits(~ j)), 2 ^ i + j ≤ n→i(# int2bits(~ j)), 2 ^ i + j < n→i(# int2bits(~ j)), j + 1 < 2 ^ i, j < 2 ^ i, j < 0 ⊦ 2 ^ n→i(# int2bits(~ j)) = 2 ^ (n→i(# uint2bits(2 ^ n→i(# int2bits(~ j)) + j)) - 1)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: int2bits-pos rotate: F subst: [i] → [~ j]

Additional information

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