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Proof step 19 from proof uint2int-rest (previous step)

uint2bits(i) = false ' + true ' + bits, 2 ^ (2 + n→i(# bits)) = 1 + bits2uint(true ' + false ' + bitsinvert(bits)) + i, 0 < i ⊦ 2 ^ (1 + n→i(# bits)) = 1 + bits2uint(bitsinvert(bits)) + i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ # bits0 ≠ 0 → bits2uint(bits + bits0) = bits2uint(bits) * 2 ^ n→i(# bits0) + bits2uint(bits0)
Proof step 21:
#(false ' + bitsinvert(bits)) ≠ 0, uint2bits(i) = false ' + true ' + bits, 2 ^ (2 + n→i(# bits)) = 1 + (bits2uint(true ') * 2 ^ n→i(#(false ' + bitsinvert(bits))) + bits2uint(false ' + bitsinvert(bits))) + i, 0 < i ⊦ 2 ^ (1 + n→i(# bits)) = 1 + bits2uint(bitsinvert(bits)) + i

Rule argument

name: bits2uint-split rotate: F subst: [bits0, bits] → [false ' + bitsinvert(bits), true ']

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ # bitsinvert(bits) = # bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + 1 ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: false ' + bitsinvert(bits), true '