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Proof step 6 from proof uint2bits-sgn-first (previous step)

bitsinvert(uint2bits(j)) +1 = boolvar + bits0 ∧ bitsinvert(uint2bits(j)) +1 ≠ [], sgn(bitsinvert(uint2bits(j)) +1) = false ' + bits, 0 < j ⊦ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
sgn(false ' + bits0) = false ' + bits, bitsinvert(uint2bits(j)) +1 = false ' + bits0, 0 < j ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ∈ sgn(boolvar0 ' + bits) ↔ boolvar ∈ boolvar0 ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ boolvar + bits = boolvar ' + bits;