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Proof step 10 from proof sgn-length-02 (previous step)

2 + # bits0 ≤ # sgn(bits0) → bits0 = [] ⊦ 2 + #(boolvar + bits0) ≤ # sgn(boolvar + bits0) → boolvar + bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# sgn(bits0) < 2 + # bits0, 3 + # bits0 ≤ # sgn(boolvar ' + bits0) ⊦ bits0 = []

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ m < n → n ≠ 0; (forward)% ⊦ bits + [] = bits;% ⊦ # [] = 0;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;