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Proof step 88 from proof byte-law-and-one (previous step)

∃ bits0, bits1. int2bits(j) = bits0 + bits1 ∧ # bits1 = 8, 8 = # fillsgn(8, int2bits(i)), # int2bits(i) < 8 ⊦ lastn(8, fillsgn(8, bitsand(int2bits(i), int2bits(j)))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 89:
8 = # fillsgn(8, int2bits(i)), # bits = 8, int2bits(j) = bits0 + bits, # int2bits(i) < 8 ⊦ bitsand(fillsgn(8, int2bits(i)), bits0 + bits) = [], lastn(8, fillsgn(8, bitsand(int2bits(i), bits0 + bits))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), bits0 + bits)))

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n = n0 → m < n0; (forward)