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Proof step 102 from proof inv-get-inverse-succ (previous step)

stutf(nσ) ∧ ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → (∃ m0. m0 ≤ m ∧ nσ[m0] = n), # nσ = ⌜m⌝ + ⌜1⌝ , stutf(nσ), n ≤ nσ[m], ∞.n < m, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = n) ∨ (∀ n0. m < n0 ∨ nσ[n0] ≠ n) ⊦ nσ[∞.n + 1] = n + 1, m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 103:
# nσ = ⌜m⌝ + ⌜1⌝ , m0 ≤ m, ∞.n < m, nσ[m0] ≤ nσ[m], stutf(nσ), (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[m0]) ∨ (∀ n0. m < n0 ∨ nσ[n0] ≠ nσ[m0]) ⊦ m = 0, nσ[∞.n + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf