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Proof step 7 from proof monls-ls (previous step)

⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, mon<(nσ), nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n0, m0, nσ0. m1 ≠ n0 - m0 ∨ ¬ m0 < n0 ∨ ¬ ⌜n0⌝ < # nσ0 ∨ ¬ mon<(nσ0) ∨ ¬ nσ0[n0] ≤ nσ0[m0] ⊦

User applied rule all left

And obtained the following new goals

Proof step 8:
⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, mon<(nσ), nσ[(m1 + 1) + m] ≤ nσ[m], m1 ≠ m1 + m - m ∨ ¬ m < m1 + m ∨ ¬ ⌜m1 + m⌝ < # nσ ∨ ¬ mon<(nσ) ∨ ¬ nσ[m1 + m] ≤ nσ[m] ⊦

Rule argument

pos: left 4 terms: m1 + m, m, nσ abort: F discard: T computed: F

Additional information

Used terms: m1 + m, m, nσ