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Proof step 4 from proof le-add-05 (previous step)

0 < i2, i + i0 + i1 + i2 ≤ i3, 0 ≤ i ⊦ i1 + i2 + i0 ≤ i + i3

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 < i2, 0 ≤ i, i + i0 + i1 + i2 ≤ i3, (i1 + i2 + i0) + i ≤ i3 ∧ i ≥ 0 → i1 + i2 + i0 ≤ i3 ⊦ i1 + i2 + i0 ≤ i + i3

Rule argument

spec: <toplevel> name: le-add-02 seq: ⊦ i + j0 ≤ i0 ∧ j0 ≥ 0 → i ≤ i0 subst: [i, j0, i0] → [i1 + i2 + i0, i, i3]

Additional information

Used terms: i1 + i2 + i0, i, i3