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Proof step 4 from proof abs-nodes-unique-strong-h (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
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
Proof step 10:
x0 = nd + x, 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

Rule argument

var: x0

Additional information

Used simplifier rules: list freely generated by [], +;;