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Proof step 4 from proof ple-01 (previous step)

i1 + 1 < j, ∀ j, i0, j0. i1 < j → i1 + i0 < j + j0 ∨ j0 < i0 ⊦ j0 < i0, (i1 + 1) + i0 < j + j0

User applied rule all left

And obtained the following new goals

Proof step 5:
i1 + 1 < j, i1 < j → i1 + 1 + i0 < j + j0 ∨ j0 < 1 + i0 ⊦ j0 < i0, (i1 + 1) + i0 < j + j0

Rule argument

pos: left 2 terms: j, 1 + i0, j0 abort: F discard: T computed: T

Additional information

Used terms: j, 1 + i0, j0