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Proof step 12 from proof sub-less (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 13:
⊦ j > 0 → i < i + j
Proof step 14:
i1 > 0 → i < i + i1, i1 < i + i1, 0 < i, j < i, 0 < j, 1 < i, 0 < i1 ⊦ j < i ∧ i < i + i1, j < i + i1

Rule argument

spec: int-abs inst: name: int-add-same-pos seq: ⊦ j > 0 → i < i + j subst: [j, i] → [i1, i] flags: current, []

Additional information

Used terms: i1, i