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Proof step 12 from proof shiftright-shift (previous step)

j + j0 = 0, 0 < j, i < 0, 0 ≤ j0, ∀ 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ 0 ≤ i ∧ 0 ≤ j → (i + j = 0 ↔ i = 0 ∧ j = 0)

Rule argument

seq: ⊦ 0 ≤ i ∧ 0 ≤ j → (i + j = 0 ↔ i = 0 ∧ j = 0) subst: [j, i] → [j0, j] rotate: F spec: int-abs inst: name: zero-eq

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used terms: j0, j