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Proof step 11 from proof valid-eraseqd-append (previous step)

valid-eraseq(q0), m ∈ eq-entry(m0, lref0) ' + q0, ∀ m1, lref1. valid-eraseq(q0 + eq-entry(m1, lref1) ') ↔ ¬ m1 ∈ q0 ⊦ ¬ m0 ∈ q0 + eq-entry(m, lref) ' ∧ valid-eraseq(q0 + eq-entry(m, lref) ') ↔ ¬ m ∈ eq-entry(m0, lref0) ' + q0 ∧ ¬ m0 ∈ q0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
valid-eraseq(q0), valid-eraseq(q0 + eq-entry(m, lref) '), m ∈ q0, ∀ m1, lref1. valid-eraseq(q0 + eq-entry(m1, lref1) ') ↔ ¬ m1 ∈ q0 ⊦ m0 ∈ q0

Rule argument

Additional information

Used simplifier rules: ⊦ n ≠ n0 → (n ∈ eq-entry(n0, lref) ' + q ↔ n ∈ q); (s)%a: ⊦ (q + q0) + q1 = q + q0 + q1; from specification list[erasequeue-base]% ⊦ n ∈ q + eq-entry(n0, lref) ' ↔ n ∈ q ∨ n = n0; (ss)