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Proof step 4 from proof even (previous step)

2 + n0 + n0 = m + m, ∀ m. n0 + n0 = m + m → n0 = m ⊦ n0 + 1 = m

User applied rule all left

And obtained the following new goals

Proof step 5:
2 + n0 + n0 = m + m, n0 + n0 = m -1 + m -1 → n0 = m -1 ⊦ n0 + 1 = m

Rule argument

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

Additional information

Used terms: m -1