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Proof step 2 from proof prefix-append-one-sizeeq (previous step)

# x0 = # x1 + 1, x0 ⊑ x1 + a ' ⊦ x0 = x1 + a '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ x ⊑ y ↔ (∃ z. x + z = y)
Proof step 4:
# x0 = # x1 + 1, ∃ z. x0 + z = x1 + a ' ⊦ x0 = x1 + a '

Rule argument

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

Additional information

Used terms: x1 + a ', x0