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Proof step 6 from proof fill-last (previous step)

⊦ bits = [], fillsgn(1, bits + boolvar ') = fillsgn(0, bits) + boolvar '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ n ≤ # bits → fillsgn(n, bits) = bits
Proof step 8:
1 ≤ #(bits + boolvar ') ⊦ bits = [], bits + boolvar ' = fillsgn(0, bits) + boolvar '

Rule argument

name: fillsgn-longer rotate: F subst: [bits, n] → [bits + boolvar ', 1]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits + boolvar ', 1