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Proof step 7 from proof int2bits-exneg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
uint2bits(~ i) = false ' + true ' + bits, false ∈ uint2bits(~ i), i < 0, i < ~1 ⊦ bits = [], ∃ bits. sgn(bitsinvert(uint2bits(~ i)) +1) = true ' + false ' + bits

Rule argument

Additional information

Used simplifier rules: uint2bits-onein-01: ⊦ j < 0 → true ∈ uint2bits(~ j);% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)