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Proof step 10 from proof valid-eraseqd-append (previous step)

valid-eraseq(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 cut applied rule cut formula

And obtained the following new goals

Proof step 11:
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
Proof step 13:
valid-eraseq(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, m ∈ eq-entry(m0, lref0) ' + q0

Rule argument

fma: m ∈ eq-entry(m0, lref0) ' + q0

Additional information

Used formulas: m ∈ eq-entry(m0, lref0) ' + q0