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Proof step 2 from proof adrlist-append-eq-size1 (previous step)

lax0 + lax1 = lax2 + ladr ', # lax2 + 1 < # lax0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
lax0 + lax1 = lax2 + ladr ', # lax2 + 1 < # lax0, ladr ∈ lax0 ∨ ladr ∈ lax1 ⊦ lax0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ ladr ∈ lax + lax0 ↔ ladr ∈ lax ∨ ladr ∈ lax0; (s)% ⊦ # lax = 0 ↔ lax = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]%in-15: ⊦ lax + ladr ' = lax0 → ladr ∈ lax0; (forward) from specification list[laddress-list]