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Proof step 7 from proof short-law-righthift-not (previous step)

⊦ ( i = 32768 ∧ j = 1 → i2s ((# int2bits(32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(32768))) )) ≠ i2s ((# int2bits(~32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~32768))) ))) ∧ ( i = 32768 ∧ j = 1 → i2s(bits2int(sublist(0, # fillsgn(32, int2bits(32768)), mklist(false, i→n(band(1, 31))) + fillsgn(32, int2bits(32768))))) ≠ i2s (bits2int (sublist (0, # fillsgn(32, int2bits(~32768)), mklist(false, i→n(bits2int(bitsand(int2bits(1), int2bits(31))))) + fillsgn(32, int2bits(~32768))))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ ( i = 32768 ∧ j = 1 → i2s ((# int2bits(32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(32768))) )) ≠ i2s ((# int2bits(~32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~32768))) ))) ∧ ( i = 32768 ∧ j = 1 → i2s (bits2int (sublist(0, # fillsgn(32, int2bits(32768)), mklist(false, i→n(bits2int(bitsand(int2bits(1), int2bits(31))))) + fillsgn(32, int2bits(32768))))) ≠ i2s (bits2int (sublist (0, # fillsgn(32, int2bits(~32768)), mklist(false, i→n(bits2int(bitsand(int2bits(1), int2bits(31))))) + fillsgn(32, int2bits(~32768))))))

Rule argument

Additional information

Used simplifier rules: ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ band((* i), (* j)) = bits2int(bitsand(int2bits(i), int2bits(j)));