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Proof step 8 from proof monls-notsmall (previous step)

nσ[0] = 0, ⌜2⌝ + ⌜n0⌝ < # nσ, mon<(nσ), nσ[2 + n0] ≤ n0 + 1, ∀ nσ0. nσ0[0] = 0 ∧ ⌜n0⌝ + ⌜1⌝ < # nσ0 ∧ mon<(nσ0) → n0 < nσ0[n0 + 1] ⊦

User applied rule all left

And obtained the following new goals

Proof step 9:
nσ[0] = 0, ⌜2⌝ + ⌜n0⌝ < # nσ, mon<(nσ), nσ[2 + n0] ≤ n0 + 1, nσ[0] = 0 ∧ ⌜n0⌝ + ⌜1⌝ < # nσ ∧ mon<(nσ) → n0 < nσ[n0 + 1] ⊦

Rule argument

pos: left 5 terms: nσ abort: F discard: T computed: T

Additional information

Used terms: nσ