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Proof step 9 from proof vals-at (previous step)

(fv ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .vals)[n0] = val0 ∨ fis0 = [] ⊦ (fv .2 ' + fis0 .vals)[n] = val

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
fv = sk1 × val1, (fv ' + fis0)[n] = sk × val, n ≤ # fis0, ∀ n0, sk0, val0. fis0[n0] = sk0 × val0 ∧ ¬ # fis0 ≤ n0 → (fis0 .vals)[n0] = val0 ∨ fis0 = [] ⊦ (val1 ' + fis0 .vals)[n] = val

Rule argument

[x0lemmaarg: spec-name: pair instance-name: fieldinit lemma-name: elim antecedent: succedent: sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val substlist: [fv, sk, val] → [fv, sk1, val1] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ sk = fv .1 ∧ val = fv .2 ↔ fv = sk × val])], [EXTRATERMS : (fv, sk1, val1)]]