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Proof step 15 from proof abs-nodes-unique-strong-h (previous step)

lax0.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x, lax0, false) ∨ lax0 ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m0 + # buf1, x, lax0.tail, false), ∀ buf2, m, x0, lax0, x1, lax1, boolvar1. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0, lax0, false) → (abs-nodes(buf2, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf1 + buf0, m, x1, lax1, boolvar1) ↔ nd ' + x = x1 ∧ lax0 = lax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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