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Proof step 12 from proof int2uint-neg (previous step)

uint2bits(~ j) = false ' + true ' + bits, false ∈ false ' + true ' + bits, j < 0 ⊦ bits2uint(sgn(bitsinvert(false ' + true ' + bits) +1)) = 2 ^ n→i(# sgn(bitsinvert(false ' + true ' + bits) +1)) + j, true ∈ (false ' + true ' + bits).rest.rest

User applied rule simplifier

And obtained the following new goals

Proof step 13:
uint2bits(~ j) = false ' + true ' + bits, j < 0 ⊦ 2 ^ n→i(# bits) = 2 ^ (1 + n→i(# bits)) + j, true ∈ bits

Rule argument

Additional information

Used simplifier rules: bits2uint-2pot: ⊦ ¬ true ∈ bits → bits2uint(true ' + bits) = 2 ^ n→i(# bits);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ ¬ boolvar ∈ bits → sgn(boolvar ' + bits) = boolvar ' + bits;% ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ false ' +1 = true ';% ⊦ ¬ true ∈ bits0 → (bits + bitsinvert(bits0))+1 = bits +1 + bits0;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bits + false ' + bits0)+1 = bits + (false ' + bits0)+1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);