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Proof step 10 from proof add-in (previous step)

(i1 + 1) + j = (i1 + 1) + i0, j + 1 = i0 + 1, ∀ j, i0. i1 + j = i1 + i0 → j = i0 ⊦ j = i0

User applied rule insert spec-lemma

And obtained the following new goals

Rule argument

spec: int-basic1 inst: name: succ-in seq: ⊦ i +1 = j +1 ↔ i = j subst: [i, j] → [j, i0] flags: []

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; Used terms: j, i0