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Proof step 36 from proof inv-inverts (previous step)

# nσ = ⌜n1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. n0 < n1 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , n1 = m0 + 1, 1 + n < n1, stutf(nσ) ⊦ ((λ m. max∞(λ n0. n0 < n1 ∧ nσ[n0] = m).n) to nσ[m0] + 1)[nσ[n]] = n, n1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# nσ = ⌜m0⌝ + ⌜1⌝ , nσ[n + 1] = nσ[n] + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = nσ[n]) = ⌜n⌝ , n < m0, stutf(nσ) ⊦ ((λ m. max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = m).n) to nσ[m0] + 1)[nσ[n]] = n

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)