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Proof step 8 from proof ladrlist-to-adr-split (previous step)

n ≤ # lax0, ∀ n, m, m0, m1, m2. n < # lax0 → (lax0.to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax2, lax1. lax0 = lax2 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax2 = n)) ∨ lax0 = [] ⊦ (@(m, m3, m4) ' + lax0.to-adr(m))[n] = @(m0, m1, m2) ↔ (∃ lax, lax1. @(m3, m4) ' + lax0 = lax + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax = n)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
(@(m, m3, m4) ' + lax0.to-adr(m))[n] = @(m0, m1, m2), n ≤ # lax0, ∀ n, m, m0, m1, m2. n < # lax0 → (lax0.to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax2, lax1. lax0 = lax2 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax2 = n)) ∨ lax0 = [] ⊦ ∃ lax, lax1. @(m3, m4) ' + lax0 = lax + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax = n
Proof step 17:
n ≤ # lax0, ∀ n, m, m0, m1, m2. n < # lax0 → (lax0.to-adr(m)[n] = @(m0, m1, m2) ↔ (∃ lax2, lax1. lax0 = lax2 + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax2 = n)) ∨ lax0 = [], ∃ lax, lax1. @(m3, m4) ' + lax0 = lax + @(m1, m2) ' + lax1 ∧ m0 = m ∧ # lax = n ⊦ (@(m, m3, m4) ' + lax0.to-adr(m))[n] = @(m0, m1, m2)

Rule argument

pos: right 1

Proof step 8 from proof ladrlist-to-adr-split (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information