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Proof step 16 from proof isprefix-add-prefixn-size (previous step)

# x0 + # z = # x1 + # x2, # x2 < n, # x1 + n < # x0, ∀ n0. n0 < # x1 + # x2 → (x0 + z)[n0] = (x1 + x2)[n0] ⊦ x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0)
Proof step 18:
# x0 + # z = # x1 + # x2, # x2 < n, ∀ n0. n0 < # x1 + # x2 → (x0 + z)[n0] = (x1 + x2)[n0], ∃ m0. (# x1 + n) + m0 = # x0 ∧ m0 ≠ 0 ⊦ x0 = []

Rule argument

spec: nat-basic name: addex seq: ⊦ m < n ↔ (∃ m0. m + m0 = n ∧ m0 ≠ 0) subst: [n, m] → [# x0, # x1 + n] flags: <all paths>

Additional information

Used terms: # x0, # x1 + n