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Proof step 22 from proof ladrlist-to-adr-split (previous step)

∀ n, m0, m3, m4, m5. ¬ # lax3 + # lax < n → ( (lax3.to-adr(m0) + @(m0, m1, m2) ' + lax.to-adr(m0))[n] = @(m3, m4, m5) ↔ (∃ lax2, lax1. lax3 + @(m1, m2) ' + lax = lax2 + @(m4, m5) ' + lax1 ∧ m3 = m0 ∧ # lax2 = n)) ⊦ (lax3.to-adr(m) + @(m, m1, m2) ' + lax.to-adr(m))[# lax3] = @(m, m1, m2), ∃ lax0, lax1. lax3 + @(m1, m2) ' + lax = lax0 + @(m1, m2) ' + lax1 ∧ # lax0 = # lax3

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: lax3, lax

Additional information

Used simplifier rules: a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list] Used terms: lax3, lax