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Proof step 33 from proof intshiftrsign32 (previous step)

# int2bits(i) = i→n(band(j, 31)) + n2, 32 = n1 + # int2bits(i), 32 = (n2 + n1) + i→n(band(j, 31)), i→n(band(j, 31)) ≤ # int2bits(i), # int2bits(i) ≤ 32, band(j, 31) ≤ 32 ⊦ bits2int(mklist(¬ 0 ≤ i, i→n(band(j, 31))) + sublist(0, # int2bits(i) - i→n(band(j, 31)), int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n2, int2bits(i))), band(j, 31) = 0, n2 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 34:
# int2bits(i) = n0 + i→n(band(j, 31)), # int2bits(i) = i→n(band(j, 31)) + n2, 32 = n1 + # int2bits(i), 32 = (n2 + n1) + i→n(band(j, 31)), i→n(band(j, 31)) ≤ # int2bits(i), # int2bits(i) ≤ 32, band(j, 31) ≤ 32 ⊦ bits2int(mklist(¬ 0 ≤ i, i→n(band(j, 31))) + sublist(0, n0, int2bits(i))) = bits2int((¬ 0 ≤ i)' + sublist(0, n2, int2bits(i))), band(j, 31) = 0, n2 = 0

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# int2bits(i), i→n(band(j, 31)), n] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # int2bits(i), i→n(band(j, 31)), n