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

i < j ⊦ i + i0 < j + j0, j0 < i0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
0 < j ⊦ j0 < i0, 0 + i0 < j + j0
Proof step 9:
i1 +1 < j, ∀ j, i0, j0. i1 < j → i1 + i0 < j + j0 ∨ j0 < i0 ⊦ j0 < i0, i1 +1 + i0 < j + j0

Rule argument

indvar: i

Additional information

Used simplifier rules: sub-pred: ⊦ i -1 = i - 1;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0;%succ-02: ⊦ i < j + 1 ↔ i ≤ j;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%sub-add-01: ⊦ i + (j - i0) = i + j - i0;%int generated by 0, +1, -1;;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ¬ i < j ∧ ¬ i0 < j0 → ¬ i + i0 < j + j0; (s)