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Proof step 15 from proof stutf-alltaken (previous step)

nσ[0] = 0, nσ[m0 + 1] = nσ[m0] + 1, ⌜m0⌝ + ⌜1⌝ < # nσ, ⌜m0⌝ < # nσ, m0 ≤ nσ[m0] → m ≤ nσ[m] ∧ nσ[m] = m0 ⊦ nσ = [], ∃ m. m ≤ m0 + 1 ∧ nσ[m] = nσ[m0] + 1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: m0 + 1

Additional information

Used simplifier rules: lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: m0 + 1