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Proof step 4 from proof odd-mult (
previous step
)
⊦ ∃ m0. 2 * m0 = m * n, 2 * m0 ≠ m ∧ 2 * m1 ≠ n
User applied rule case distinction
And obtained the following new goals
Proof step 5
:
⊦ 2 * m0 ≠ m, ∃ m0. 2 * m0 = m * n
Proof step 7
:
⊦ 2 * m1 ≠ n, ∃ m0. 2 * m0 = m * n
Rule argument
pos: right 2
Additional information