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Proof step 3 from proof isprefix-app-size-ls1 (previous step)

# x1 + # x2 < # x0, x0 ⊑ x1 + x2 ⊦ x0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 5:
# x1 + # x2 < # x0, ∃ z. x0 + z = x1 + x2 ⊦ x0 = []

Rule argument

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

Additional information

Used terms: x1 + x2, x0