| ∃ i. i + i = j0 ∨ i + i + 1 = j0 ⊦ ∃ i. i + i = j0 -1 ∨ i + i + 1 = j0 -1 |
i + i = j0 ∨ i + i + 1 = j0 ⊦ ∃ i0. (i0 + i0) + 1 = j0 ∨ 2 + i0 + i0 = j0
Used simplifier rules: succ-01: ⊦ i +1 = i + 1;%add-a: ⊦ i + j + i0 = (i + j) + i0;%add-c: ⊦ i + j = j + i;%pred: ⊦ i = j -1 ↔ i +1 = j; from specification int-basic1