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Proof step 10 from proof sub-less (previous step)

i1 < i + i1, 1 < i, 0 < i1, 0 < j, j < i, 0 < i ⊦ j < i + i1

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 11:
⊦ i < j ∧ j < j0 → i < j0
Proof step 12:
0 < i, j < i, 0 < j, 0 < i1, 1 < i, i1 < i + i1, j < i ∧ i < i + i1 → j < i + i1 ⊦ j < i + i1

Rule argument

spec: int-basic1 name: trans seq: ⊦ i < j ∧ j < j0 → i < j0 subst: [i, j, j0] → [j, i, i + i1]

Additional information

Used terms: j, i, i + i1