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Proof step 7 from proof shiftright-shift (previous step)

j + j0 < 32, 0 ≤ j, 0 ≤ j0, i < 0, ∀ j0, i, j1. i→n(j1) < i→n(j) → i < 0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j1 + j0 < 32 → (i >> j1) >> j0 = i >> (j1 + j0) ⊦ (i >> j) >> j0 = i >> (j + j0)

User applied rule cut formula

And obtained the following new goals

Proof step 8:
j + j0 < 32, 0 ≤ j, 0 ≤ j0, i < 0, ∀ j0, i, j1. i→n(j1) < i→n(j) → i < 0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j1 + j0 < 32 → (i >> j1) >> j0 = i >> (j1 + j0) ⊦ j = 0, (i >> j) >> j0 = i >> (j + j0)

Rule argument

fma: j = 0

Additional information

Used simplifier rules: rightshift-zero: ⊦ i >> 0 = i;% ⊦ i + 0 = i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used formulas: j = 0