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Proof step 3 from proof stutf-notbig (previous step)

⌜n0 +1⌝ < # nσ, stutf(nσ), ∀ nσ. ⌜n0⌝ < # nσ ∧ stutf(nσ) → (n0 < nσ[n0] ↔ false) ⊦ n0 +1 < nσ[n0 +1] ↔ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⌜n0⌝ + ⌜1⌝ < # nσ, n0 + 1 < nσ[n0 + 1], stutf(nσ), ∀ nσ0. ⌜n0⌝ < # nσ0 ∧ stutf(nσ0) → ¬ n0 < nσ0[n0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%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)