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Proof step 5 from proof addex (previous step)

∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0) ⊦ n0 = 0, n0 < m, ∃ m0. m + m0 = n0 + 1 ∧ m0 ≠ 0

User applied rule all left

And obtained the following new goals

Proof step 6:
m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0), ∀ m. m < n0 → n0 = 0 ∨ (∃ m0. m + m0 = n0 ∧ m0 ≠ 0) ⊦ n0 = 0, n0 < m, ∃ m0. m + m0 = n0 + 1 ∧ m0 ≠ 0

Rule argument

pos: left 1 terms: m abort: F discard: F computed: F

Additional information

Used terms: m