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Proof step 7 from proof mon-lem (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 8:
⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, mon(σσ), m1 = m1 + m - m ∧ ⌜m1 + m⌝ < # σσ ∧ mon(σσ) ∧ m ≤ m1 + m → (∃ σ1. σσ[m] + σ1 = σσ[m1 + m]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

pos: left 3 terms: m1 + m, m, σσ abort: F discard: T computed: F

Additional information

Used terms: m1 + m, m, σσ