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Proof step 166 from proof byte-law-and-one (previous step)

true ' + fillsgn(7, int2bits(i)) = fillsgn(8, int2bits(i)), # int2bits(i) ≤ # bits1 ∧ int2bits(i) ≠ [] ∧ int2bits(i).first, # bits2 + # bits1 = 8, int2bits(j) = (bits0 + bits2) + bits1, # bits1 = # int2bits(i), 7 = # fillsgn(7, int2bits(i)), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < 8, i < 0, # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i))) = [], lastn(8, fillsgn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i)))), bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], bits2 = []

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

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