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Proof step 3 from proof monls-ls (previous step)

∀ n, m, nσ. ¬(m1 = n - m ∧ nσ[n] ≤ nσ[m] ∧ ⌜n⌝ < # nσ ∧ m < n ∧ mon<(nσ)), m1 +1 = n - m, nσ[n] ≤ nσ[m], ⌜n⌝ < # nσ, m < n, mon<(nσ) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m1 + 1 = n - m, nσ[n] ≤ nσ[m], ⌜n⌝ < # nσ, m < n, mon<(nσ), ∀ n0, m0, nσ0. m1 ≠ n0 - m0 ∨ ¬ nσ0[n0] ≤ nσ0[m0] ∨ ¬ ⌜n0⌝ < # nσ0 ∨ ¬ m0 < n0 ∨ ¬ mon<(nσ0) ⊦

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2