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Proof step 3 from proof prefix-app-size-eq (previous step)

# x0 = # x1, x0 ⊑ x1 + x2 ⊦ x0 = x1

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used simplifier rules: ⊦ # x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0); (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1 + x2, x0