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Proof step 5 from proof nls (previous step)

i + i0 < j + j0, j < i ⊦ i0 < j0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
j < i, i + i0 < j + j0 ⊦ i0 = j0, i0 < j0

Rule argument

fma: i0 = j0

Additional information

Used simplifier rules: ls-01: ⊦ i + j < i + i0 ↔ j < i0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;% ⊦ i < j → ¬ j < i; (s) Used formulas: i0 = j0