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Proof step 2 from proof le-add-05 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i < j ↔ ¬ j ≤ i
Proof step 4:
0 < i2, i + i0 + i1 + i2 ≤ i3, 0 ≤ i ⊦ i1 + i2 + i0 ≤ i + i3

Rule argument

spec: <toplevel> name: lels seq: ⊦ i < j ↔ ¬ j ≤ i subst: [j, i] → [i1 + i2 + i0, i + i3] flags: <all paths>

Additional information

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