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Proof step 2 from proof isprefix-eq-size-app-eq1 (previous step)

# x0 = # x1 + # x2 ⊦ x1 + x2 ⊑ x0 ↔ x0 = x1 + x2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# x0 = # x1 + # x2, x1 + x2 ⊑ x0 ⊦ x0 = x1 + x2

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x ⊑ x; (s)% ⊦ x + y ⊑ x + z ↔ y ⊑ z; (s)