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Proof step 2 from proof le-add-07 (
previous step
)
i + i1 < j + i2, 0 < i0, 0 ≤ i, i0 + i2 + j ≤ i1 ⊦
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 3
:
⊦ i < j ↔ ¬ j ≤ i
Proof step 4
:
0 < i0, i0 + i2 + j ≤ i1, 0 ≤ i ⊦ j + i2 ≤ i + i1
Rule argument
spec: <toplevel> name: lels seq: ⊦ i < j ↔ ¬ j ≤ i subst: [j, i] → [j + i2, i + i1] flags: <all paths>
Additional information
Used terms: j + i2, i + i1