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Proof step 5 from proof square-even (
previous step
)
i * i + i = 2 * j + 1, i = 2 * j0 ∨ i = 2 * j1 + 1 ⊦
User applied rule case distinction
And obtained the following new goals
Proof step 6
:
i = 2 * j0, i * i + i = 2 * j + 1 ⊦
Proof step 9
:
i = 2 * j1 + 1, i * i + i = 2 * j + 1 ⊦
Rule argument
pos: left 2
Additional information