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Proof step 3 from proof even-add (previous step)

⊦ (i + j) % 2 = 0 ↔ even(i) ∧ even(j) ∨ odd(i) ∧ odd(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ (i + j) % 2 = 0 ↔ i % 2 = 0 ∧ j % 2 = 0 ∨ i % 2 ≠ 0 ∧ j % 2 ≠ 0
Proof step 5:
⊦ i % 2 = 0 ∧ j % 2 = 0 ∨ i % 2 ≠ 0 ∧ j % 2 ≠ 0 ↔ even(i) ∧ even(j) ∨ odd(i) ∧ odd(j)

Rule argument

spec: int-div inst: name: even-add seq: ⊦ (i + j) % 2 = 0 ↔ i % 2 = 0 ∧ j % 2 = 0 ∨ i % 2 ≠ 0 ∧ j % 2 ≠ 0 subst: [j, i] → [j, i] flags: rewrite

Additional information

Used terms: j, i