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Proof step 1 from proof ladr-split (conclusion)

⊦ lax.to-adr(n) = ax0 + ax1 ↔ (∃ lax0, lax1. lax = lax0 + lax1 ∧ ax0 = lax0.to-adr(n) ∧ ax1 = lax1.to-adr(n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
lax.to-adr(n) = ax0 + ax1 ⊦ ∃ lax0, lax1. lax = lax0 + lax1 ∧ ax0 = lax0.to-adr(n) ∧ ax1 = lax1.to-adr(n)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ (lax0 + lax1).to-adr(n) = lax0.to-adr(n) + lax1.to-adr(n); (s)