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Proof step 4 from proof lst-02 (previous step)

i < j + i1 + 1, ∀ i, j, j0. i < j + i1 → i < j0 + i1 ∨ j0 < j ⊦ j0 < j, i < j0 + i1 + 1

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: i, j + 1, j0 + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%ls-01: ⊦ i + j < i + i0 ↔ j < i0; Used terms: i, j + 1, j0 + 1