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Proof step 25 from proof fill-last (previous step)

(boolvar0 ' + bits0) + boolvar ' ≠ [] ∧ #((boolvar0 ' + bits0) + boolvar ') < 2 + m0, # bits0 < m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ ((boolvar0 ' + bits0) + boolvar ').first ' + fillsgn((2 + m0)-1, (boolvar0 ' + bits0) + boolvar ') = boolvar0 ' + fillsgn(m0, boolvar0 ' + bits0) + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# bits0 < m0, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ fillsgn(1 + m0, (boolvar0 ' + bits0) + boolvar ') = fillsgn(m0, boolvar0 ' + bits0) + boolvar '

Rule argument

Additional information

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