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Proof step 12 from proof split-prefix-append-to (
previous step
)
# σ0 = ∞, # σ1 = ∞, ∀ n1. ((σ0 to n) + σ1)[n1] = σ0[n1] ⊦ σ1[n0] = σ0[n + n0]
User applied rule all left
And obtained the following new goals
Proof step 13
:
# σ1 = ∞, # σ0 = ∞, ((σ0 to n) + σ1)[n + n0] = σ0[n + n0] ⊦ σ1[n0] = σ0[n + n0]
Rule argument
pos: left 3 terms: n + n0 abort: F discard: T computed: T
Additional information
Used terms: n + n0