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Proof step 7 from proof monle-leq (previous step)

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

User applied rule all left

And obtained the following new goals

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