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Proof step 18 from proof stutf-alltaken (previous step)

⌜m0⌝ + ⌜1⌝ < # nσ, stutf(nσ), m ≤ m0, nσ[m] ≤ nσ[m0 + 1], ∀ nσ, n. ⌜m0⌝ < # nσ ∧ stutf(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: ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m