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Proof step 5 from proof abs-nodes-unique-strong-h (previous step)

x0 = [], abs-nodes(buf, m, x0, lax0, false), ∀ buf0, m, x0, lax0, x1, lax1, boolvar1. # buf0 < # buf → abs-nodes(buf0, m, x0, lax0, false) → (abs-nodes(buf0, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
abs-nodes(buf, m, [], [], false), ∀ buf0, m, x0, lax0, x1, lax1, boolvar1. # buf0 < # buf → abs-nodes(buf0, m, x0, lax0, false) → (abs-nodes(buf0, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1) ⊦ abs-nodes(buf, m, x1, lax1, boolvar1) ↔ [] = x1 ∧ [] = lax1

Rule argument

Additional information

Used simplifier rules: abs-nodes-elems-empty-adr-empty-h: ⊦ abs-nodes(buf, m, [], lax, weak) → lax = [];