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Proof step 51 from proof inv-inverts-leq (previous step)

mon≤(nσ) ∧ m + 1 ≤ n0 ∧ ⌜n0⌝ < # nσ → nσ[m + 1] ≤ nσ[n0], nσ[n1] = nσ[n], nσ[n0] = nσ[n], # nσ = ∞, m + 1 < n1, m + 1 < n0, stutf(nσ), nσ[n] ≤ nσ[m], ∞.n < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
# nσ = ∞, nσ[n0] = nσ[n], nσ[n1] = nσ[n], ∞.n < n, nσ[n] ≤ nσ[m], stutf(nσ), m + 1 < n0, m + 1 < n1, m < n0, nσ[m + 1] ≤ nσ[n], nσ[m + 1] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

Additional information

Used simplifier rules: stutf-monle: ⊦ stutf(nσ) → (¬ mon≤(nσ) ↔ false);%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2%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