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Proof step 2 from proof lst-01 (previous step)

i < j + i0, j < j0, j0 + i0 < j + i0 ⊦ i < j0 + i0

Heuristic structural induction applied rule structural induction

And obtained the following new goals

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

Rule argument

term: i0

Additional information

Used simplifier rules: add-zero: ⊦ i + 0 = i;%add-a: ⊦ i + j + i0 = (i + j) + i0;%int generated by 0, +1, -1;;%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1