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Proof step 13 from proof isprefix-add-prefixn-size (previous step)

x0 + z = x1 + x2, # x2 < n, # x1 + n < # x0 ⊦ x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n])
Proof step 15:
#(x0 + z) = #(x1 + x2), # x1 + n < # x0, # x2 < n, ∀ n. n < #(x0 + z) → (x0 + z)[n] = (x1 + x2)[n] ⊦ x0 = []

Rule argument

spec: list-del name: extension seq: ⊦ x = y ↔ # x = # y ∧ (∀ n. n < # x → x[n] = y[n]) subst: [y, x] → [x1 + x2, x0 + z] flags: <all paths>

Additional information

Used terms: x1 + x2, x0 + z