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

n < m0, 2 + m0 + m0 < n + n, ∀ n0. m0 + m0 < n0 + n0 ∧ n0 < m0 → n0 + n0 = 0 ∨ m0 = n0 ⊦ n + n = 0, m0 + 1 = n, m0 = 0, m0 = n

User applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 3 terms: n -1 abort: F discard: T computed: F

Additional information

Used terms: n -1