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Proof step 283 from proof byte-law-and-one (previous step)

8 ≤ #(bits2 + bitsand(bits1, bits)), int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, bits.first ⊦ lastn(8, bitsand(bits1, bits)) = lastn(8, bits2 + bitsand(bits1, bits)), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 284:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, bits.first, 8 ≤ # bits2 + # bitsand(bits1, bits) ⊦ lastn(8, bitsand(bits1, bits)) = lastn(8, bits2 + bitsand(bits1, bits)), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = [], # bits2 + # bitsand(bits1, bits) = 0

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;