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Proof step 3 from proof payload-one (previous step)

m < # zbrar ⊦ payload(zbrar, m, 1) = (zbrar[m].entry? ⊃ (∅)[zbrar[m].key, zbrar[m].adr];∅)

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ m < # zbrar → payload(zbrar, m, nused + 1) = (zbrar[m].entry? ⊃ payload(zbrar, m + 1, nused)[zbrar[m].key, zbrar[m].adr];payload(zbrar, m + 1, nused))

Rule argument

spec: <toplevel> name: payload-rec seq: ⊦ m < # zbrar → payload(zbrar, m, nused + 1) = (zbrar[m].entry? ⊃ payload(zbrar, m + 1, nused)[zbrar[m].key, zbrar[m].adr] ;payload(zbrar, m + 1, nused) ) subst: [m, zbrar, nused] → [m, zbrar, 0]

Additional information

Used simplifier rules: payload-base: ⊦ payload(zbrar, m, 0) = ∅;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: m, zbrar, 0