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Proof step 1 from proof even-add (conclusion)

⊦ even(i + j) ↔ even(i) ∧ even(j) ∨ odd(i) ∧ odd(j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ even(i) ↔ i % 2 = 0
Proof step 3:
⊦ (i + j) % 2 = 0 ↔ even(i) ∧ even(j) ∨ odd(i) ∧ odd(j)

Rule argument

spec: inst: name: even-def-mod seq: ⊦ even(i) ↔ i % 2 = 0 subst: [i] → [i + j] flags: rewrite, current

Additional information

Used terms: i + j