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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], ∀ n1. ((σ0 to n) + σ1)[n1] = σ0[n1] ⊦ σ1[n0] = σ0[n + n0]

Rule argument

pos: left 3 terms: n + n0 abort: F discard: F computed: T

Additional information

Used terms: n + n0