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Proof step 21 from proof int2bits-falsein-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
uint2bits(~ i) = (false ' + bits1) + true ' + bits0, bitsinvert(uint2bits(~ i)) +1 = (true ' + bitsinvert(bits1)) + true ' + bits0, i < 0, i < ~1, false ∈ uint2bits(~ i) ⊦ bits1 = [], true ∈ bits0, false ∈ sgn((true ' + bitsinvert(bits1)) + true ' + bits0).rest, (bitsinvert(bits1) + true ').first

Rule argument

Additional information

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