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Proof step 7 from proof bits2uint-ones (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j
Proof step 9:
0 ≤ n→i(# bits0), ∀ bits1. # bits1 < #(bits0 + true ') → true → bits1 = [] ∨ 2 ^ n→i(# bits1) = bits2uint(bits1) + 1 ∨ false ∈ bits1 ⊦ 2 ^ n→i(# bits0) * 2 = 2 * (bits2uint(bits0) + 1), false ∈ bits0

Rule argument

seq: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j subst: [i, j] → [n→i(# bits0), 2] rotate: F spec: int-pot inst: name: succ

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n); Used terms: n→i(# bits0), 2