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Proof step 23 from proof fill-last (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits)
Proof step 25:
(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 '

Rule argument

name: fill-rec rotate: F subst: [bits, n] → [(boolvar0 ' + bits0) + boolvar ', 2 + m0]

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ ¬ n < n; Used terms: (boolvar0 ' + bits0) + boolvar ', 2 + m0