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Proof step 56 from proof int2bits-falsein-01 (previous step)

¬(¬ false ∈ bitsinvert(bits1) ∧ 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, true ∈ bits1 ⊦ bits1 = [], true ∈ bits0, false ∈ sgn(bitsinvert(bits1)).rest, false ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
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)).rest, false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ bitsinvert(bits) = [] ↔ bits = [];% ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits;