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Proof step 26 from proof int2bits-neg-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j
Proof step 28:
0 ≤ i, ~1 = 2 ^ i * 2 + j, n→i(# int2bits(~ j)) = i + 1, 0 ≤ 2 ^ i, j + 1 < 2 ^ i, 0 < 2 ^ i + j, j < 0, j < 2 ^ i, 1 < 2 ^ i, 0 < 2 ^ i ⊦

Rule argument

seq: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j subst: [i, j] → [i, 2] rotate: F spec: int-pot inst: name: succ

Additional information

Used simplifier rules: ⊦ 0 < i → (0 < i ^ j ↔ i = 1 ∨ 0 ≤ j); Used terms: i, 2