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Proof step 17 from proof abs-nodes-elems-adr-length-h (previous step)

lax.head = @(m0, m1), abs-nodes(buf0, m + # buf1, nd ' + x0, lax, weak), abs-padding(buf1), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ # x0 + 1 = # lax

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
abs-nodes(buf0, m + # buf1, nd ' + x0, @(m0, m1) ' + lax0, weak), abs-padding(buf1), ∀ buf2, m, x, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, weak) → # x = # lax ⊦ # x0 = # lax0

Rule argument

Additional information

Used simplifier rules: abs-nodes-first-adr-explicit: ⊦ abs-nodes(buf, m, a ' + x, lax, weak) → (lax.head = ladr ↔ (∃ lax0. lax = ladr + lax0));% ⊦ ladr ∈ ladr0 ' + lax ↔ ladr = ladr0 ∨ ladr ∈ lax; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m +1 = m + 1; (s)% ⊦ #(ladr ' + lax) = (# lax)+1; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]%in-14: ⊦ ladr ' + lax = lax0 → ladr ∈ lax0; (forward) from specification list[laddress-list]% ⊦ ladr + lax = ladr ' + lax; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list