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Proof step 23 from proof min-skolemf (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 24:
mon<(nσ), ⌜m0⌝ < # nσ ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ[m0] ≤ nσ[m] → m0 ≤ m) ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < nσ[m0], ∀ n0, nσ0. n0 < nσ[m0] +1 → mon<(nσ0) → nσ0 = [] ∨ (∃ m0. ⌜m0⌝ < # nσ0 ∧ n0 ≤ nσ0[m0] ∧ (∀ m. ⌜m⌝ < # nσ0 ∧ n0 ≤ nσ0[m] → m0 ≤ m)) ∨ # nσ0 ≠ ∞ ∧ nσ0[(# nσ0).n - 1] < n0 ⊦ nσ = [], ∃ m. ⌜m⌝ < # nσ ∧ nσ[m0] < nσ[m] ∧ (∀ m1. ⌜m1⌝ < # nσ ∧ nσ[m0] < nσ[m1] → m ≤ m1), # nσ ≠ ∞ ∧ ¬ nσ[m0] < nσ[(# nσ).n - 1], ⌜m0 + 1⌝ < # nσ ∧ nσ[m0] < nσ[m0 + 1] ∧ (∀ m1. ⌜m1⌝ < # nσ ∧ nσ[m0] < nσ[m1] → m0 + 1 ≤ m1)

Rule argument

pos: right 2 terms: m0 + 1 abort: F discard: F computed: F

Additional information

Used terms: m0 + 1