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Proof step 40 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 ⊦ bits1 = [], true ∈ bits0, false ∈ sgn(true ' + true ' + bits0).rest, true ∈ bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
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 ∈ bits0, true ∈ bits1

Rule argument

Additional information

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