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Proof step 48 from proof eq-leninv (previous step)

# nσ = ∞, m + 1 = nσ[n], stutf(nσ), ∀ m0. n < m0 → nσ[m0] = m + 1 ⊦ m + 1 = 0, ⌜m + 1⌝ = leninv(nσ)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 49:
⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝
Proof step 50:
# nσ = ∞, m + 1 = nσ[n], stutf(nσ), ∀ m0. n < m0 → nσ[m0] = m + 1 ⊦ ⌜m⌝ + ⌜1⌝ = leninv(nσ), m + 1 = 0

Rule argument

spec: nat-inf name: add-fin seq: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ subst: [n, m] → [1, m] flags: <all paths>

Additional information

Used terms: 1, m