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Proof step 36 from proof eq-leninv (previous step)

# nσ = ∞, nσ[m2] + 1 = nσ[n], n < m2, m0 ≤ m2, stutf(nσ), n < m0, nσ[n] < nσ[m0], ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = nσ[n], m0 = 0, nσ[m0] = 0, m2 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 37:
⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n]

Rule argument

spec: inst: name: monle-leq seq: ⊦ mon≤(nσ) ∧ m ≤ n ∧ ⌜n⌝ < # nσ → nσ[m] ≤ nσ[n] subst: [nσ, m, n] → [nσ, m0, m2] flags: current, []

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: nσ, m0, m2