[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 14 from proof and-011x-neg (previous step)

∀ bits, bits2, bits3. # bits < # bits1 → bitsand(false ' + bits, true ' + bits2) = false ' + bits3 → 2 ^ n→i(# bits) = 1 + bits2uint(bitsinvert(bits2)) % 2 ^ n→i(# bits) + bits2uint(bits3) ∨ false ∈ bits ⊦ 2 ^ n→i(# bits1) = 1 + bits2uint(bits1), false ∈ bits1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2uint(bits) = 2 ^ n→i(# bits) - 1

Rule argument

seq: ⊦ ¬ false ∈ bits ∧ bits ≠ [] → bits2uint(bits) = 2 ^ n→i(# bits) - 1 subst: [bits] → [bits1] rotate: F spec: int2bits inst: name: bits2uint-ones

Additional information

Used simplifier rules: ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ # [] = 0;% ⊦ bits2uint([]) = 0; Used terms: bits1