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Proof step 11 from proof mon-lem (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 12:
σσ[m] + σ = σσ[m1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ⌜m1 + m⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[m1 + m] + σ1 = σσ[(m1 + m) + 1]), ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

pos: left 4 terms: m1 + m abort: F discard: F computed: F

Additional information

Used terms: m1 + m