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Proof step 4 from proof getkeys-app (previous step)

∀ fis0. (fis1 + fis0) .keys = fis1 .keys + fis0 .keys ⊦ (((sk × val) ' + fis1) + fis0) .keys = ((sk × val) ' + fis1) .keys + fis0 .keys

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ((sk × val) ' + fis) .keys = sk ' + fis .keys
Proof step 3:
∀ fis0. (fis1 + fis0) .keys = fis1 .keys + fis0 .keys ⊦ (fis1 + fis0) .keys = fis1 .keys + fis0 .keys

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + sks0 = sks + sks1 ↔ sks0 = sks1], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]