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Proof step 11 from proof int2bits-exneg (previous step)

false ∈ false ' + true ' + bits, i < 0, i < ~1 ⊦ bits = [], sgn(bitsinvert(false ' + true ' + bits) +1) = true ' + false ' + sgn(true ' + (false ' + bitsinvert(bits))+1).rest.rest, ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
i < 0, i < ~1 ⊦ bits = [], sgn(true ' + (false ' + bitsinvert(bits))+1) = true ' + false ' + sgn(true ' + (false ' + bitsinvert(bits))+1).rest.rest, ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

Rule argument

Additional information

Used simplifier rules: ⊦ 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);