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Proof step 22 from proof abs-nodes-aligned-headersize-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) → is-aligned(# buf2, NODE_HEADER_SIZE) ⊦ is-aligned(# buf1 + # buf0, NODE_HEADER_SIZE)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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) → is-aligned(# buf2, NODE_HEADER_SIZE) ⊦ is-aligned(# buf1 + # buf0, NODE_HEADER_SIZE)

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)%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