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Proof step 9 from proof ex-ls-add (previous step)

∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 = j ∨ i0 < j ⊦ ∃ i. j + 1 = i0 + i ∧ 0 < i

User applied rule cut formula

And obtained the following new goals

Proof step 10:
∀ j. i0 < j → (∃ i1. j = i0 + i1 ∧ 0 < i1), i0 = j ∨ i0 < j ⊦ i0 = j, ∃ i. j + 1 = i0 + i ∧ 0 < i

Rule argument

fma: i0 = j

Additional information

Used simplifier rules: add-in: ⊦ i + j = i + i0 ↔ j = i0;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i; Used formulas: i0 = j