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Proof step 30 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 ∈ (true ' + bitsinvert(bits1) + true ' + bits0).rest, bitsinvert(bits1).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
uint2bits(~ i) = (false ' + bits1) + true ' + bits0, bitsinvert(uint2bits(~ i)) +1 = (true ' + bitsinvert(bits1)) + true ' + bits0, i < 0, i < ~1, false ∈ uint2bits(~ i) ⊦ bits1 = [], bitsinvert(bits1).first, (bitsinvert(bits1) + true ').first, true ∈ bits0, false ∈ bits0, true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;% ⊦ false ∈ bitsinvert(bits) ↔ true ∈ bits;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (boolvar ' + bits).rest = bits;% ⊦ bits + boolvar ' + bits0 ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ' + bits0).first ↔ (bits + boolvar ').first;