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Proof step 5 from proof modls-03 (previous step)

n1 < n, n2 < n, m0 + 1 < m, m * n + n2 ≤ (n * m0 + n) + n1, ∀ n0, n3, m1, n4. ¬ n0 < n3 ∨ ¬ m0 < m1 ∨ ¬ n4 < n3 ∨ ¬ m1 * n3 + n0 ≤ m0 * n3 + n4 ⊦

User applied rule all left

And obtained the following new goals

Proof step 6:
m0 + 1 < m, n2 < n, n1 < n, m * n + n2 ≤ (n * m0 + n) + n1, ¬ n2 < n ∨ ¬ m0 < m - 1 ∨ ¬ n1 < n ∨ ¬ (m - 1) * n + n2 ≤ m0 * n + n1 ⊦

Rule argument

pos: left 5 terms: n2, n, m - 1, n1 abort: F discard: T computed: F

Additional information

Used terms: n2, n, m - 1, n1