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Proof step 34 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
# nσ = ∞, nσ[m2] + 1 = nσ[n], nσ[n] < nσ[m0], n < m0, stutf(nσ), ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = nσ[n]

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (mon≤(nσ) ↔ true);% ⊦ m < n → n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ∞; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → m ≤ n; (s)