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Proof step 4 from proof isprefix-add-prefixn-size (previous step)

# x1 + n < # x0, x0 ⊑ x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2) ⊦ x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 6:
# x1 + n < # x0, ∃ z. x0 + z = x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2) ⊦ x0 = []

Rule argument

spec: list-last name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2), x0] flags: <all paths>

Additional information

Used terms: x1 + (n ≤ # x2 ⊃ firstn(n, x2);x2), x0