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Proof step 7 from proof intshiftrsign-if (previous step)

# int2bits(i) ≤ i→n(band(j, 31)) ⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = (¬ 0 ≤ i ⊃ ~1;0)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 8:
¬ 0 ≤ i, # int2bits(i) ≤ i→n(band(j, 31)) ⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = ~1
Proof step 11:
0 ≤ i, # int2bits(i) ≤ i→n(band(j, 31)) ⊦ bits2int(sublist(0, # int2bits(i), mklist(¬ 0 ≤ i, i→n(band(j, 31))) + int2bits(i))) = 0

Rule argument

pos: right 1

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