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Proof step 16 from proof mon-lem (previous step)

σσ[m] + σ = σσ[m1 + m], σσ[m1 + m] + σ0 = σσ[m1 + 1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ σσ[m] + σ + σ0 = σσ[m1 + 1 + m]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ m + n = n + m
Proof step 18:
σσ[m] + σ = σσ[m1 + m], σσ[m1 + m] + σ0 = σσ[m1 + m + 1], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ σσ[m] + σ + σ0 = σσ[m1 + m + 1]

Rule argument

spec: nat-basic2 inst: name: c seq: ⊦ m + n = n + m subst: [n, m] → [m, 1] flags: rewrite

Additional information

Used terms: m, 1