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Proof step 988 from proof bitsand-cast (previous step)

# bits = # bits1 ⊦ bits1 = [], bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits2, bits3) + bits1)), bits3 + bits = [], bits2 + bits1 = [], bits3 = [], bits2 = [], true ∈ bits1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 989:
⊦ n ≤ # bits → fillsgn(n, bits) = bits
Proof step 990:
# bits1 ≤ #(bitsand(bits2, bits3) + bits1), # bits = # bits1 ⊦ bits1 = [], bits = [], bits1 = lastn(# bits1, bitsand(bits2, bits3) + bits1), bits3 + bits = [], bits2 + bits1 = [], bits3 = [], bits2 = [], true ∈ bits1

Rule argument

seq: ⊦ n ≤ # bits → fillsgn(n, bits) = bits subst: [bits, n] → [bitsand(bits2, bits3) + bits1, # bits1] rotate: F spec: bitopsaux inst: name: fillsgn-longer

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (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: bitsand(bits2, bits3) + bits1, # bits1