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Proof step 20 from proof inv-inverts-leq (previous step)

# nσ = ∞, stutf(nσ), ∞.n < n, ∃ m0. n < m0 ∧ nσ[n] < nσ[m0], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# nσ = ∞, ∞.n < m, nσ[n] < nσ[m], n < m, ∞.n < n, stutf(nσ), ∞.n + 1 < m, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%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%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1