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Proof step 33 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%lels-03: ⊦ m < n → m ≤ n; from specification nat