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

i + i0 < j + j0 ⊦ i + i1 < j + j1, i0 < i1, j1 < j0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i + i0 < j + j0 ⊦ j1 = j0, i + i1 < j + j1, i0 < i1, j1 < j0

Rule argument

fma: j1 = j0

Additional information

Used simplifier rules: add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; Used formulas: j1 = j0