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Proof step 30 from proof in-payload (previous step)

key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].entry? ∧ zbrar[(m + 1) + n].key = key), m + nused0 < # zbrar, m < # zbrar ⊦ key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. ¬ nused0 < n ∧ zbrar[m + n].entry? ∧ zbrar[m + n].key = key), zbrar[m].entry?

User applied rule case distinction

And obtained the following new goals

Proof step 31:
key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].entry? ∧ zbrar[(m + 1) + n].key = key), m < # zbrar, m + nused0 < # zbrar, key ∈ payload(zbrar, m + 1, nused0) ⊦ zbrar[m].entry?, ∃ n. ¬ nused0 < n ∧ zbrar[m + n].entry? ∧ zbrar[m + n].key = key
Proof step 35:
key ∈ payload(zbrar, m + 1, nused0) ↔ (∃ n. n < nused0 ∧ zbrar[(m + 1) + n].entry? ∧ zbrar[(m + 1) + n].key = key), m < # zbrar, m + nused0 < # zbrar, ∃ n. ¬ nused0 < n ∧ zbrar[m + n].entry? ∧ zbrar[m + n].key = key ⊦ zbrar[m].entry?, key ∈ payload(zbrar, m + 1, nused0)

Rule argument

pos: right 1

Proof step 30 from proof in-payload (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information