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Proof step 43 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, m < m0, nσ[m] < nσ[m0], stutf(nσ), stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ nσ[m] + 1 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = nσ[m] + 1) ⊦ ∃ m0. nσ[m0] = nσ[m] + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
# nσ = ∞, nσ[m1] = nσ[m] + 1, m < m0, nσ[m] < nσ[m0], stutf(nσ), m1 ≤ m0, nσ[m1] ≤ nσ[m0] ⊦ ∃ m0. nσ[m0] = nσ[m1]

Rule argument

Additional information

Used simplifier rules: eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ ⌜m⌝ < ∞; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)