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Proof step 7 from proof even-06 (previous step)

3 = 2 + 1, 2 + m0 + m0 = 3 + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1, ∀ n. m0 + m0 ≠ 3 + n + n ∨ ¬(∀ m. m + m ≠ n + n + 1) ⊦

User applied rule insert equation

And obtained the following new goals

Proof step 8:
2 + m0 + m0 = (2 + 1) + n0 + n0, ∀ m. m + m ≠ n0 + n0 + 1, ∀ n. m0 + m0 ≠ (2 + 1) + n + n ∨ ¬(∀ m. m + m ≠ n + n + 1) ⊦

Rule argument

pos: left 1 flags: discard

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