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Proof step 4 from proof add-in (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
(i1 + 1) + j = (i1 + 1) + i0, i1 + 1 + j = i1 + 1 + i0 → 1 + j = 1 + i0, ∀ j, i0. i1 + j = i1 + i0 → j = i0 ⊦ j = i0

Rule argument

pos: left 2 terms: 1 + j, 1 + i0 abort: F discard: F computed: F

Additional information

Used terms: 1 + j, 1 + i0