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Proof step 5 from proof add-even (previous step)

j1 % 2 = 0 ∧ j2 % 2 = 0 ∨ j1 % 2 ≠ 0 ∧ j2 % 2 ≠ 0 ⊦ (j1 ^2 + j2 ^2) % 2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ (i + j) % 2 = 0 ↔ i % 2 = 0 ∧ j % 2 = 0 ∨ i % 2 ≠ 0 ∧ j % 2 ≠ 0
Proof step 7:
j1 % 2 = 0 ∧ j2 % 2 = 0 ∨ j1 % 2 ≠ 0 ∧ j2 % 2 ≠ 0 ⊦ j1 ^2 % 2 = 0 ∧ j2 ^2 % 2 = 0, j1 ^2 % 2 ≠ 0 ∧ j2 ^2 % 2 ≠ 0

Rule argument

spec: int-div name: even-add seq: ⊦ (i + j) % 2 = 0 ↔ i % 2 = 0 ∧ j % 2 = 0 ∨ i % 2 ≠ 0 ∧ j % 2 ≠ 0 subst: [j, i] → [j2 ^2, j1 ^2] flags: <all paths>

Additional information

Used terms: j2 ^2, j1 ^2