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Proof step 2 from proof mod2-add-neg (previous step)

j % 2 = 0, abs(i) ≤ j, i ≤ 0 ⊦ i % 2 + (i + j) % 2 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
0 < j, j % 2 = 0, abs(i) ≤ j, i ≤ 0 ⊦ i % 2 + (i + j) % 2 = 0

Rule argument

fma: 0 < j

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%abs-zero: ⊦ abs(i) ≤ 0 ↔ i = 0; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: 0 < j