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Proof step 21 from proof mininv-inverts-inner-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
nσ[n] < nσ[n + 1], nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜min(nσ, nσ[n] + 1)⌝ < # nσ, ⌜n⌝ < # nσ, ⌜n⌝ + ⌜1⌝ < # nσ, mon<(nσ), min(nσ, nσ[n] + 1) ≤ n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1] ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = []

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)