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Proof step 5 from proof square-even (
previous step
)
i * i + i = 2 * j + 1, (∃ j. i = 2 * j) ∨ (∃ j. i = 2 * j + 1) ⊦
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
i * i + i = 2 * j + 1, i = 2 * j0 ∨ i = 2 * j1 + 1 ⊦
Rule argument
Additional information