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Proof step 5 from proof three (previous step)

m + n0 + 1 = 3, ∀ m. m + n0 = 3 → m = 1 ∨ n0 = 0 ∨ m = 0 ∨ m = 2 ∧ n0 = 1 ⊦ m = 0, m = 1, m = 0, m = 2 ∧ n0 = 0

User applied rule all left

And obtained the following new goals

Proof step 6:
m + n0 + 1 = 3, m +1 + n0 = 3 → m +1 = 1 ∨ n0 = 0 ∨ m +1 = 0 ∨ m +1 = 2 ∧ n0 = 1 ⊦ m = 0, m = 1, m = 0, m = 2 ∧ n0 = 0

Rule argument

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

Additional information

Used terms: m +1