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Proof step 7 from proof q-add-01 (previous step)

m0 + m = n ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ ∃ m. (m0 + 1) + m = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

User applied rule exists right

And obtained the following new goals

Proof step 8:
m0 + m = n ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ (m0 + 1) + m -1 = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Rule argument

pos: right 1 terms: m -1 abort: F discard: T computed: F

Additional information

Used terms: m -1