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Proof step 36 from proof abs-nodes-unique-weak-h (previous step)

lax0.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x, lax0, true) ∨ lax0 ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m + # buf1, x, lax0.tail, true), #(buf1 + buf0) = 0 ∧ [] = [] ∨ (∃ buf3, buf2. buf1 + buf0 = buf3 + buf2 ∧ abs-padding(buf3) ∧ abs-nodes(buf2, m + # buf3, [], [], true)) ∨ true ∧ abs-failed(buf1 + buf0) ∧ [] = [], ∀ buf2, m, x0, lax0, x1, lax1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, true) ∧ abs-nodes(buf2, m, x1, lax1, true) → x0 = x1 ∧ lax0 = lax1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
lax0.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x, lax0, true) ∨ lax0 ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m + # buf1, x, lax0.tail, true), # buf1 + # buf0 = 0 ∨ buf1 + buf0 = buf + buf2 ∧ abs-padding(buf) ∧ abs-nodes(buf2, m + # buf, [], [], true) ∨ abs-failed(buf1 + buf0), ∀ buf2, m, x0, lax0, x1, lax1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, true) ∧ abs-nodes(buf2, m, x1, lax1, true) → x0 = x1 ∧ lax0 = lax1 ⊦

Rule argument

Additional information

Used simplifier rules: append-a: ⊦ (buf + buf0) + buf1 = buf + buf0 + buf1; from specification array-splice[buffer]% ⊦ #(buf + buf0) = # buf + # buf0; (s)