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Proof step 2 from proof plst (previous step)

i < j ⊦ i + i0 < j + j0, j0 < i0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i < j ⊦ j0 = i0, i + i0 < j + j0, j0 < i0

Rule argument

fma: j0 = i0

Additional information

Used simplifier rules: ls-add-both: ⊦ i < j → i1 + i < i1 + j;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i; Used formulas: j0 = i0