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Proof step 15 from proof fill-last (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ n ≤ # bits → fillsgn(n, bits) = bits
Proof step 17:
2 + m0 ≤ #(bits + boolvar '), m0 < # bits, ∀ bits, boolvar. bits = [] ∨ fillsgn(m0 + 1, bits + boolvar ') = fillsgn(m0, bits) + boolvar ' ⊦ bits + boolvar ' = fillsgn(m0 + 1, bits) + boolvar ', bits = []

Rule argument

name: fillsgn-longer rotate: F subst: [bits, n] → [bits + boolvar ', 2 + m0]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ ¬ 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: bits + boolvar ', 2 + m0