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Proof step 20 from proof band-bitlength (previous step)

# int2bits(j) < n, # int2bits(j) < # int2bits(i), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) < n, n ≤ # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(i) < # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(j) < # sgn(bitsand(int2bits(i), int2bits(j))), n ≤ # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < n, # int2bits(j) + 1 < # sgn(bitsand(int2bits(i), int2bits(j))), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), j < 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits

Rule argument

name: bitsand-len-true rotate: F subst: [bits0, bits] → [int2bits(j), int2bits(i)]

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ j < 0 → int2bits(j).first;% ⊦ m < n → m ≤ n;% ⊦ ¬ n < n; Used terms: int2bits(j), int2bits(i)