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Proof step 3 from proof mon-lem (previous step)

⌜n⌝ < # σσ, mon(σσ), m ≤ n ⊦ ∃ σ1. σσ[m] + σ1 = σσ[n]

User applied rule structural induction

And obtained the following new goals

Proof step 4:
m1 +1 = n - m, ⌜n⌝ < # σσ, mon(σσ), m ≤ n, ∀ n, m, σσ. m1 = n - m ∧ ⌜n⌝ < # σσ ∧ mon(σσ) ∧ m ≤ n → (∃ σ1. σσ[m] + σ1 = σσ[n]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[n]

Rule argument

term: n - m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;% ⊦ m ≤ n → (n - m = 0 ↔ n = m); (s)% ⊦ m ≤ m; (s)% ⊦ σ0 + σ = σ0 ↔ σ = [] ∨ ¬ finp(σ0); (s)%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream