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Proof step 103 from proof rightshift-8-01 (previous step)

127 < i ∧ i ≤ 255, i→n(j) < #(i2b(0) ' + i2b(i - 256) '), 0 ≤ i, i ≤ 255, ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ i / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(#(i2b(0) ' + i2b(i - 256) ') - i→n(j), i2b(0) ' + i2b(i - 256) ')), j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 104:
0 < i, i→n(j) < 2, 127 < i, i ≤ 255, 0 ≤ i, ∀ i, j0. i→n(j0) < i→n(j) → i→n(j0) < # int2bytes(i) → i >> 8 * abs(j0) = bytes2int(firstn(# int2bytes(i) - i→n(j0), int2bytes(i))) ⊦ i / 256 >> 8 * (abs(j) - 1) = bytes2int(firstn(2 - i→n(j), i2b(0) ' + i2b(i - 256) ')), j = 0

Rule argument

Additional information

Used simplifier rules: i0 ≠ 0, i0 < i ⊦ (* i) < j → (* i0) < j;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;