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Proof step 9 from proof abs-nodes-prefix-elim (previous step)

abs-nodes(buf0 + buf1, 0, x1, lax1, boolvar1), abs-nodes(buf0, 0, x0, lax0, false), lax0 ⊑ lax1, ∃ z. x0 + z = x1 ⊦ ∃ x2, lax2. x1 = x0 + x2 ∧ lax1 = lax0 + lax2 ∧ abs-nodes(buf1, # buf0, x2, lax2, boolvar1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
abs-nodes(buf0, 0, x0, lax0, false), abs-nodes(buf0 + buf1, 0, x0 + z, lax1, boolvar1), lax0 ⊑ lax1 ⊦ ∃ lax. lax1 = lax0 + lax ∧ abs-nodes(buf1, # buf0, z, lax, boolvar1)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s)