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Proof step 60 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 61:
nσ[m0 + 1] = nσ[n], nσ[m0] + 1 = nσ[n], # nσ = ∞, m0 < m, nσ[m0] < nσ[n], m0 < n, stutf(nσ), nσ[n] < nσ[m], nσ[m0] < nσ[m], ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ m = 0, nσ[m] = nσ[n], n = 0, nσ[n] = 0, nσ[m] = 0

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%lels-03: ⊦ m < n → m ≤ n; from specification nat%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat