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Proof step 33 from proof path-equiv-reach (previous step)

x = a + y0 ∧ x ≠ [], x .last = y .butlast .last, # x = n0 + 1, # y = 1 + n0, path(x), path(a ' + y), reach(a, x .last, n0) ⊦ y = [], ∃ c. reach(a, c, n0) ∧ y .last ∈ succ(c)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
(a ' + y0) .last = y .butlast .last, # y = 1 + # y0, reach(a, y .butlast .last, # y0), path(a ' + y), path(a ' + y0) ⊦ y = [], ∃ c. reach(a, c, # y0) ∧ y .last ∈ succ(c)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ a ' + x ≠ []], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ (x + y) + z = x + y + z], [CSIMPSEQ : ⊦ # a ' = 0 +1], [CSIMPSEQ : ⊦ # (x + y) = # x + # y], [CSIMPSEQ : ⊦ a + x = a ' + x])]]