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Proof step 6 from proof monls-ls (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
mon<(nσ), nσ[(m1 + 1) + m] ≤ nσ[m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, m1 ≠ m1 + m - m ∨ ¬ nσ[m1 + m] ≤ nσ[m] ∨ ¬ ⌜m1 + m⌝ < # nσ ∨ ¬ m < m1 + m ∨ ¬ mon<(nσ) ⊦

Rule argument

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

Additional information

Used terms: m1 + m, m, nσ