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Proof step 3 from proof two-01 (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 4:
m0 + n +1 = 2 → n +1 = 0 ∨ m0 = 0 ∨ m0 = 1 ∧ n +1 = 1, m0 +1 + n = 2 ⊦ n = 0, m0 +1 = 0, m0 +1 = 1 ∧ n = 1

Rule argument

pos: left 1 terms: n +1 abort: F discard: T computed: F

Additional information

Used terms: n +1