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Proof step 17 from proof stutf-alltaken (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: m

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m