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Proof step 39 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)) ⊦ fillsgn(n, bitsand(fillsgn(n, true '), bits0)) = fillsgn(n, bits0), bits0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 40:
⊦ 0 < n → fillsgn(n, boolvar ') = mklist(boolvar, n)
Proof step 41:
0 < n, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = []

Rule argument

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

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ fillsgn(0, bits) = bits; Used terms: true, n