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Proof step 4 from proof even-02 (previous step)

2 + m0 + m0 < n + n, ∀ n. m0 + m0 < n + n → n + n = 0 ∨ m0 < n ⊦ n + n = 0, m0 + 1 < n

User applied rule all left

And obtained the following new goals

Proof step 5:
2 + m0 + m0 < n + n, m0 + m0 < n -1 + n -1 → n -1 + n -1 = 0 ∨ m0 < n -1 ⊦ n + n = 0, m0 + 1 < n

Rule argument

pos: left 2 terms: n -1 abort: F discard: T computed: F

Additional information

Used terms: n -1