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Proof step 21 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], fillsgn(n, bitsand(fillsgn(n, false '), bits0)) = fillsgn(n, false ')

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n)
Proof step 23:
0 < n, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], fillsgn(n, bitsand(mklist(false, n), bits0)) = mklist(false, n)

Rule argument

seq: ⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n) subst: [boolvar, n] → [false, n] rotate: F spec: bitopsaux inst: name: fill-rec-one

Additional information

Used simplifier rules: and-zero-02: ⊦ bits ≠ [] → bitsand(false ', bits) = false ';% ⊦ 0 < n ↔ n ≠ 0;% ⊦ fillsgn(0, bits) = bits;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0; Used terms: false, n