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Proof step 3 from proof valid-eraseq-strong-notinvols (previous step)

n < # apebs, valid-eraseq-strong(apebs, vtbl, q), valid-peb(v, m, apebs[n]), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ valid-eraseq-strong(apebs, vtbl, q + eq-entry(n, v × m) '), n ∈ q

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 4:
n < # apebs, valid-eraseq-strong(apebs, vtbl, []), valid-peb(v, m, apebs[n]), ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ valid-eraseq-strong(apebs, vtbl, [] + eq-entry(n, v × m) '), n ∈ []
Proof step 15:
n < # apebs, valid-eraseq-strong(apebs, vtbl, eqent + q0), valid-peb(v, m, apebs[n]), ∀ n, apebs, vtbl, v, m. n < # apebs ∧ valid-eraseq-strong(apebs, vtbl, q0) ∧ valid-peb(v, m, apebs[n]) ∧ (¬ v ∈ vtbl ∨ ¬ m < vtbl[v]) → valid-eraseq-strong(apebs, vtbl, q0 + eq-entry(n, v × m) ') ∨ n ∈ q0, ¬ v ∈ vtbl ∨ ¬ m < vtbl[v] ⊦ valid-eraseq-strong(apebs, vtbl, (eqent + q0) + eq-entry(n, v × m) '), n ∈ eqent + q0

Rule argument

term: q

Additional information

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