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Proof step 5 from proof expand-cons (previous step)

x0 = [] ∨ (∃ a, σ0. ⌜x0⌝ = ⌜a '⌝ + σ0) ⊦ a + x0 = [], ∃ a0, σ0. ⌜a + x0⌝ = ⌜a0 '⌝ + σ0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x0 = [] ∨ ⌜x0⌝ = ⌜a0 '⌝ + σ ⊦ ∃ a1, σ0. ⌜a '⌝ + ⌜x0⌝ = ⌜a1 '⌝ + σ0

Rule argument

Additional information

Used simplifier rules: append-finfin: ⊦ ⌜x + y⌝ = ⌜x⌝ + ⌜y⌝ ;%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list