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Proof step 24 from proof ubits2bits (previous step)

(bits1 + true ').first, ∀ bits0. # bits0 < #(bits1 + true ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ (n→i(# bits1) + 1) = 2 * ((bits2uint(bits1) + 1) + bits2uint(bitsinvert(bits1)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j
Proof step 26:
0 ≤ n→i(# bits1), (bits1 + true ').first, ∀ bits0. # bits0 < #(bits1 + true ') → bits0.first → bits0 = [] ∨ 2 ^ n→i(# bits0) = bits2uint(bits0) + bits2uint(bitsinvert(bits0)) + 1 ⊦ 2 ^ n→i(# bits1) * 2 = 2 * ((bits2uint(bits1) + 1) + bits2uint(bitsinvert(bits1)))

Rule argument

seq: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j subst: [i, j] → [n→i(# bits1), 2] rotate: F spec: int-pot inst: name: succ

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n); Used terms: n→i(# bits1), 2