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Proof step 8 from proof byte-law-righthift-not (previous step)

⊦ ( i = 128 ∧ j = 1 → i2b ((# int2bits(128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(128))) )) ≠ i2b ((# int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~128))) ))) ∧ ( i = 128 ∧ j = 1 → i2b (bits2int (sublist(0, # fillsgn(32, int2bits(128)), mklist(false, i→n(bits2int(bitsand(int2bits(1), int2bits(31))))) + fillsgn(32, int2bits(128))))) ≠ i2b (bits2int (sublist(0, # fillsgn(32, int2bits(~128)), mklist(false, i→n(bits2int(bitsand(int2bits(1), int2bits(31))))) + fillsgn(32, int2bits(~128))))))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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