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Proof step 7 from proof shiftright-is-le (previous step)

i < 0, 0 < (i >> j) + i, ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ i < 0 ∧ j < 0 → ¬ 0 < i + j
Proof step 9:
¬(i >> j < 0 ∧ i < 0), i < 0, 0 < (i >> j) + i, ∀ j, i0. i→n(i0) < i→n(i) → abs(i0) < i0 >> j → false ⊦ j = 0

Rule argument

seq: ⊦ i < 0 ∧ j < 0 → ¬ 0 < i + j subst: [j, i] → [i, i >> j] rotate: F spec: int-basic2 inst: name: ls-zero

Additional information

Used terms: i, i >> j