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Proof step 2 from proof pls (previous step)

i0 < j0, i < j ⊦ i + i0 < j + j0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
i1 +1 < j, i0 < j0, ∀ i0, j0, j. i0 < j0 ∧ i1 < j → i1 + i0 < j + j0 ⊦ i1 +1 + i0 < j + j0
Proof step 5:
i1 -1 < j, i0 < j0, ∀ i0, j0, j. i0 < j0 ∧ i1 < j → i1 + i0 < j + j0 ⊦ i1 -1 + i0 < j + j0

Rule argument

indvar: i

Additional information

Used simplifier rules: add-zero: ⊦ i + 0 = i;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%ls: ⊦ i + j < j ↔ i < 0;%int generated by 0, +1, -1;;%lf-04: ⊦ ¬ j < i ∧ j < i0 → i < i0; (forward) from specification int-basic1% ⊦ i < j → ¬ j < i; (s)