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Proof step 4 from proof le-add-06 (previous step)

i0 + i + i1 ≤ i2, 0 ≤ i ⊦ i1 + i0 ≤ i + i2

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ i + j0 ≤ i0 ∧ j0 ≥ 0 → i ≤ i0
Proof step 6:
0 ≤ i, i0 + i + i1 ≤ i2, (i0 + i1) + i ≤ i2 ∧ i ≥ 0 → i0 + i1 ≤ i2 ⊦ i1 + i0 ≤ i + i2

Rule argument

spec: <toplevel> name: le-add-02 seq: ⊦ i + j0 ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 subst: [i, j0, i0] → [i0 + i1, i, i2]

Additional information

Used terms: i0 + i1, i, i2