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Proof step 188 from proof byte-law-and-one (previous step)

7 = # fillsgn(7, int2bits(i)), # bits1 = # int2bits(i), int2bits(j) = (bits0 + bits2) + bits1, # bits2 + # bits1 = 8, 8 ≤ # bits0 + # bits2 + # bitsand(bits1, int2bits(i)), 8 ≤ # bits2 + # bitsand(bits1, int2bits(i)), # bits1 < # bits2 + # bitsand(bits1, int2bits(i)), # bits1 < # bits0 + # bits2 + # bitsand(bits1, int2bits(i)), # bits1 < 8, i < 0 ⊦ bits1 = [], bits2 + bitsand(bits1, int2bits(i)) = [], lastn(8, bits0 + bits2 + bitsand(bits1, int2bits(i))) = lastn(8, bits2 + bitsand(bits1, int2bits(i))), bits2 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ n ≤ # bits → lastn(n, bits0 + bits) = lastn(n, bits) subst: [bits, bits0, n] → [bits2 + bitsand(bits1, int2bits(i)), bits0, 8] rotate: F spec: list-del inst: bitlista name: lastN-append-01

Additional information

Used simplifier rules: ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m ≤ n → (n < m ↔ false); Used terms: bits2 + bitsand(bits1, int2bits(i)), bits0, 8