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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2