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Proof step 30 from proof lookup-list-put-ex (previous step)

∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1) ⊦ (st[r0 -- sk, val][r -- (sk0 ' + sks0)]) = st[r -- (sk0 ' + sks0)], ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 9:
¬ sks0 ≠ [], ∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1) ⊦ (st[r0 -- sk, val][r -- (sk0 ' + sks0)]) = st[r -- (sk0 ' + sks0)], ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0
Proof step 29:
sks0 ≠ [], ∀ st0, r1, sk1, val0, r2. (st0[r1 -- sk1, val0][r2 -- sks0]) ≠ st0[r2 -- sks0] → (∃ sks, sks1. sks0 = sks + sk1 ' + sks1 ∧ st0[r2 -- sks] .refval = r1) ⊦ (st[r0 -- sk, val][(st[r0 -- sk, val][r -- sk0]) .refval -- sks0]) = st[r -- (sk0 ' + sks0)], ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: lookup-rec-01 antecedent: succedent: sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks] substlist: [sks, sk, r, st] → [sks0, sk0, r, st[r0 -- sk, val]] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((1, 0, 0, 1, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks0, sk0, r, st[r0 -- sk, val])]]