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Proof step 16 from proof ladrlist-to-adr-split (previous step)

(lax.to-adr(m) + @(m, m1, m2) ' + lax1.to-adr(m))[# lax] = @(m, m1, m2), ∀ n, m0, m5, m6, m7. ¬ # lax + # lax1 < n → ( (lax.to-adr(m0) + @(m0, m1, m2) ' + lax1.to-adr(m0))[n] = @(m5, m6, m7) ↔ (∃ lax2, lax3. lax + @(m1, m2) ' + lax1 = lax2 + @(m6, m7) ' + lax3 ∧ m5 = m0 ∧ # lax2 = n)) ⊦ ∃ lax2, lax3. @(m3, m4) ' + lax + @(m1, m2) ' + lax1 = lax2 + @(m1, m2) ' + lax3 ∧ # lax2 = # lax + 1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: @(m3, m4) ' + lax, lax1

Additional information

Used simplifier rules: a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ #(ladr ' + lax) = (# lax)+1; (s)% ⊦ m +1 = m + 1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: @(m3, m4) ' + lax, lax1