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Proof step 50 from proof int2bits-falsein-01 (previous step)

true ' + bits0 ≠ [] ∧ (¬ (true ' + bits0).first) ∈ bitsinvert(bits1), uint2bits(~ i) = (false ' + bits1) + true ' + bits0, bitsinvert(uint2bits(~ i)) +1 = (true ' + bitsinvert(bits1)) + true ' + bits0, i < 0, i < ~1, false ∈ uint2bits(~ i), (bitsinvert(bits1) + true ').first ⊦ bits1 = [], true ∈ bits0, false ∈ (sgn(bitsinvert(bits1)) + true ' + bits0).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
uint2bits(~ i) = (false ' + bits1) + true ' + bits0, bitsinvert(uint2bits(~ i)) +1 = (true ' + bitsinvert(bits1)) + true ' + bits0, i < 0, i < ~1, false ∈ uint2bits(~ i), (bitsinvert(bits1) + true ').first, true ∈ bits1 ⊦ bits1 = [], true ∈ bits0, false ∈ (sgn(bitsinvert(bits1)) + true ' + bits0).rest

Rule argument

Additional information

Used simplifier rules: ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];