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Proof step 13 from proof int2bits-exneg (previous step)

i < 0, i < ~1 ⊦ true ∈ bits, 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i < 0, i < ~1 ⊦ bits = [], bits = false ' + bits.rest, true ∈ bits, ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

Rule argument

Additional information

Used simplifier rules: ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ ¬ boolvar ∈ bits → sgn(boolvar ' + bits) = boolvar ' + bits;% ⊦ sgn(boolvar ' + boolvar ' + bits) = sgn(boolvar ' + bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ false ' +1 = true ';% ⊦ ¬ true ∈ bits0 → (bits + bitsinvert(bits0))+1 = bits +1 + bits0;