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Proof step 4 from proof stutf-alltaken (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⌜m0⌝ + ⌜1⌝ < # nσ, stutf(nσ), n ≤ nσ[m0 + 1], ∀ nσ, n. ⌜m0⌝ < # nσ ∧ stutf(nσ) ∧ n ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = n) ⊦ ∃ m. m ≤ m0 + 1 ∧ nσ[m] = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ m +1 = m + 1; (s)