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Proof step 6 from proof ints2bytes-butlast (previous step)

ints0 = [] ∨ is2bs(ints0) .butlast = is2bs(ints0 .butlast) ⊦ i + ints0 = [], is2bs(i + ints0) .butlast = is2bs((i + ints0) .butlast)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is2bs(i ' + ints) = i2b(i) ' + is2bs(ints)
Proof step 2:
⊦ is2bs([]) = []
Proof step 5:
is2bs(ints0) .butlast = is2bs(ints0 .butlast) ⊦ ints0 = [], (i2b(i) ' + is2bs(ints0)) .butlast = i2b(i) ' + is2bs(ints0 .butlast)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ints ≠ [] → (ints0 + ints) .butlast = ints0 + ints .butlast], [CSIMPSEQ : ⊦ i ' .butlast = []], [CSIMPSEQ : ⊦ ints + [] = ints], [CSIMPSEQ : ⊦ by ' .butlast = []], [CSIMPSEQ : ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1], [CSIMPSEQ : ⊦ bytes + [] = bytes], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ i ' + ints ≠ []], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]