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Proof step 34 from proof lastn-fill-sgn (previous step)

bits0 = boolvar + bits, # bits0 = # fillsgn(# bits0, sgn(bits0)), boolvar, true ∈ bits1, # sgn(bits0) < # bits1 + # bits0, # sgn(bits0) < # bits0, 1 + # sgn(bits0) < # bits1 + # bits0 ⊦ bits0 = [], bits1 + bits0 = [], bits1 = [], fillsgn(# bits0, sgn(bits0)) = bits0, false ∈ bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
1 + # bits = # fillsgn(1 + # bits, sgn(true ' + bits)), # sgn(true ' + bits) < # bits + # bits1, # sgn(true ' + bits) ≤ # bits, true ∈ bits1 ⊦ fillsgn(1 + # bits, sgn(true ' + bits)) = true ' + bits, bits1 = [], false ∈ bits1

Rule argument

Additional information

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