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Proof step 2 from proof firstn-prefix-01 (previous step)

x = firstn(n, x) + z ⊦ firstn(n, x) ⊑ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 4:
x = firstn(n, x) + z ⊦ ∃ z. firstn(n, x) + z = x

Rule argument

spec: list-last inst: name: prefix seq: ⊦ x ⊑ y ↔ (∃ z. x + z = y) subst: [y, x] → [x, firstn(n, x)] flags: rewrite

Additional information

Used terms: x, firstn(n, x)